আমরা জানি,
\(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B} ........(1)\)
ধরি, \(B=A\)
\((1)\) হতে,
\(\sin{(A+A)}=\sin{A}\cos{A}+\cos{A}\sin{A}\)
\(\therefore \sin{2A}=2\sin{A}\cos{A}\)
\(\sin{2A}=2\sin{A}\cos{A}\)
আবার,
\(\sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 1+\sin{2A}=1+2\sin{A}\cos{A}\) ➜Note উভয় পার্শে \(1\) যোগ করে।

\(\Rightarrow 1+\sin{2A}=\sin^2{A}+\cos^2{A}+2\sin{A}\cos{A}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)

\(\therefore 1+\sin{2A}=(\sin{A}+\cos{A})^2\) ➜Note \(\because a^2+b^2+2ab=(a+b)^2\)

\(1+\sin{2A}=(\sin{A}+\cos{A})^2\)
আবার,
\(\sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow -1+\sin{2A}=-1+2\sin{A}\cos{A}\) ➜Note উভয় পার্শে \((-1)\) যোগ করে।

\(\Rightarrow -(1-\sin{2A})=-(1-2\sin{A}\cos{A})\)
\(\Rightarrow 1-\sin{2A})=1-2\sin{A}\cos{A}\)
\(\Rightarrow 1-\sin{2A}=\sin^2{A}+\cos^2{A}-2\sin{A}\cos{A}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)

\(\therefore 1-\sin{2A}=(\sin{A}-\cos{A})^2\) ➜Note \(\because a^2+b^2-2ab=(a-b)^2\)

\(1-\sin{2A}=(\sin{A}-\cos{A})^2\)

আমরা জানি,
\(\sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow \sin{2A}=\cos^2{A}\times2\frac{\sin{A}}{\cos{A}}\)
\(=\frac{1}{\sec^2{A}}\times2\tan{A}\) ➜Note \(\because \cos{A}=\frac{1}{\sec{A}}\)
এবং \(\frac{\sin{A}}{\cos{A}}=\tan{A}\)

\(=\frac{2\tan{A}}{\sec^2{A}}\)
\(=\frac{2\tan{A}}{1+\tan^2{A}}\) ➜Note \(\because \sec^2{A}=1+\tan^2{A}\)

\(\sin{2A}=\frac{2\tan{A}}{1+\tan^2{A}}\)
\(\sin{2A}=\frac{2\tan{A}}{1+\tan^2{A}}\)

আমরা জানি,
\(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B} ........(1)\)
ধরি, \(B=A\)
\((1)\) হতে,
\(\cos{(A+A)}=\cos{A}\cos{A}-\sin{A}\sin{A}\)
\(\therefore \cos{2A}=\cos^2{A}-\sin^2{A}\)
\(\cos{2A}=\cos^2{A}-\sin^2{A}\)
আবার,
\(\cos{2A}=\cos^2{A}-\sin^2{A}\)
\(=\cos^2{A}-(1-\cos^2{A})\) ➜Note \(\because \sin^2{A}=1-\cos^2{A}\)

\(=\cos^2{A}-1+\cos^2{A}\)
\(=2\cos^2{A}-1\)
\(\therefore \cos{2A}=2\cos^2{A}-1\)
\(\cos{2A}=2\cos^2{A}-1\)
আবার,
\(\cos{2A}=\cos^2{A}-\sin^2{A}\)
\(=1-\sin^2{A}-\sin^2{A}\) ➜Note \(\because \cos^2{A}=1-\sin^2{A}\)

\(=1-2\sin^2{A}\)
\(\therefore \cos{2A}=1-2\sin^2{A}\)
\(\cos{2A}=1-2\sin^2{A}\)
আবার,
\(\cos{2A}=2\cos^2{A}-1\)
\(\therefore 1+\cos{2A}=2\cos^2{A}\) ➜Note পক্ষান্তর করে।

\(1+\cos{2A}=2\cos^2{A}\)
আবার,
\(\cos{2A}=1-2\sin^2{A}\)
\(\Rightarrow -1+\cos{2A}=-2\sin^2{A}\) ➜Note পক্ষান্তর করে।

\(\Rightarrow -(1-\cos{2A})=-2\sin^2{A}\)
\(\therefore 1-\cos{2A}=2\sin^2{A}\)
\(1-\cos{2A}=2\sin^2{A}\)

আমরা জানি,
\(\cos{2A}=\cos^2{A}-\sin^2{A}\)
\(=\cos^2{A}\left(1-\frac{\sin^2{A}}{\cos^2{A}}\right)\)
\(=\frac{1}{\sec^2{A}}\left(1-\tan^2{A}\right)\) ➜Note \(\because \cos{A}=\frac{1}{\sec{A}}\)
এবং \(\frac{\sin{A}}{\cos{A}}=\tan{A}\)

\(=\frac{1-\tan^2{A}}{\sec^2{A}}\)
\(=\frac{1-\tan^2{A}}{1+\tan^2{A}}\) ➜Note \(\because \sec^2{A}=1+\tan^2{A}\)

\(\therefore \cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)
\(\cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)
আবার,
\(1-\cos{2A}=2\sin^2{A} .......(1)\)
\(1+\cos{2A}=2\cos^2{A} .......(2)\)
\((1)\) কে \((2)\) দ্বারা ভাগ করে।
\(\frac{1-\cos{2A}}{1+\cos{2A}}=\frac{2\sin^2{A}}{2\cos^2{A}}\)
\(=\frac{\sin^2{A}}{\cos^2{A}}\)
\(=\tan^2{A}\)
\(\therefore \frac{1-\cos{2A}}{1+\cos{2A}}=\tan^2{A}\)
\(\frac{1-\cos{2A}}{1+\cos{2A}}=\tan^2{A}\)

আমরা জানি,
\(\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}} ........(1)\)
ধরি, \(B=A\)
\((1)\) হতে,
\(\tan{(A+A)}=\frac{\tan{A}+\tan{A}}{1-\tan{A}\tan{A}}\)
\(\therefore \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)
\(\tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)

আমরা জানি,
\(\cot{(A+B)}=\frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}} ........(1)\)
ধরি, \(B=A\)
\((1)\) হতে,
\(\cot{(A+A)}=\frac{\cot{A}\cot{A}-1}{\cot{A}+\cot{A}}\)
\(\therefore \cot{2A}=\frac{\cot^2{A}-1}{2\cot{A}}\)
\(\cot{2A}=\frac{\cot^2{A}-1}{2\cot{A}}\)

আমরা জানি,
\(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B} ........(1)\)
ধরি, \(B=2A\)
\((1)\) হতে,
\(\sin{(A+2A)}=\sin{A}\cos{2A}+\cos{A}\sin{2A}\)
\(\Rightarrow \sin{3A}=\sin{A}(1-2\sin^2{A})+\cos{A}(2\sin{A}\cos{A})\) ➜Note \(\because \cos{2A}=1-2\sin^2{A}\)
এবং \(\sin{2A}=2\sin{A}\cos{A}\)

\(=\sin{A}-2\sin^3{A}+2\sin{A}\cos^2{A}\)
\(=\sin{A}-2\sin^3{A}+2\sin{A}(1-\sin^2{A})\) ➜Note \(\because \cos^2{A}=1-\sin^2{A}\)

\(=\sin{A}-2\sin^3{A}+2\sin{A}-2\sin^3{A}\)
\(=3\sin{A}-4\sin^3{A}\)
\(\therefore \sin{3A}=3\sin{A}-4\sin^3{A}\)
\(\sin{3A}=3\sin{A}-4\sin^3{A}\)

আমরা জানি,
\(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B} ........(1)\)
ধরি, \(B=2A\)
\((1)\) হতে,
\(\cos{(A+2A)}=\cos{A}\cos{2A}-\sin{A}\sin{2A}\)
\(\Rightarrow \cos{3A}=\cos{A}(2\cos^2{A}-1)-\sin{A}(2\sin{A}\cos{A})\) ➜Note \(\because \cos{2A}=2\cos^2{A}-1\)
এবং \(\sin{2A}=2\sin{A}\cos{A}\)

\(=2\cos^3{A}-\cos{A}-2\cos{A}\sin^2{A}\)
\(=2\cos^3{A}-\cos{A}-2\cos{A}(1-\cos^2{A})\) ➜Note \(\because \sin^2{A}=1-\cos^2{A}\)

\(=2\cos^3{A}-\cos{A}-2\cos{A}+2\cos^3{A}\)
\(=4\cos^3{A}-3\cos{A}\)
\(\therefore \cos{3A}=4\cos^3{A}-3\cos{A}\)
\(\cos{3A}=4\cos^3{A}-3\cos{A}\)

আমরা জানি,
\(\tan{(A+B+C)}=\frac{\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C}}{1-\tan{A}\tan{B}-\tan{B}\tan{C}-\tan{C}\tan{A}} ......(1)\)
ধরি,
\(B=C=A \)
\((1)\) হতে,
\(\tan{(A+A+A)}=\frac{\tan{A}+\tan{A}+\tan{A}-\tan{A}\tan{A}\tan{A}}{1-\tan{A}\tan{A}-\tan{A}\tan{A}-\tan{A}\tan{A}}\)
\(\Rightarrow \tan{3A}=\frac{3\tan{A}-\tan^3{A}}{1-\tan^2{A}-\tan^2{A}-\tan^2{A}}\)
\(\therefore \tan{3A}=\frac{3\tan{A}-\tan^3{A}}{1-3\tan^2{A}}\)
\(\tan{3A}=\frac{3\tan{A}-\tan^3{A}}{1-3\tan^2{A}}\)

আমরা জানি,
\(\cot{(A+B+C)}=\frac{\cot{A}\cot{B}\cot{C}-\cot{A}-\cot{B}-\cot{C}}{\cot{B}\cot{C}+\cot{C}\cot{A}+\cot{A}\cot{B}-1} ......(1)\)
ধরি,
\(B=C=A \)
\((1)\) হতে,
\(\cot{(A+A+A)}=\frac{\cot{A}\cot{A}\cot{A}-\cot{A}-\cot{A}-\cot{A}}{\cot{A}\cot{A}+\cot{A}\cot{A}+\cot{A}\cot{A}-1}\)
\(\Rightarrow \cot{3A}=\frac{\cot^3{A}-3\cot{A}}{\cot^2{A}+\cot^2{A}+\cot^2{A}-1}\)
\(\therefore \cot{3A}=\frac{\cot^3{A}-3\cot{A}}{3\cot^2{A}-1}\)
\(\cot{3A}=\frac{\cot^3{A}-3\cot{A}}{3\cot^2{A}-1}\)

লেখা যায়,
\(\cot{A}-\tan{A}=\frac{\cos{A}}{\sin{A}}-\frac{\sin{A}}{\cos{A}}\)
\(=\frac{\cos^2{A}-\sin^2{A}}{\sin{A}\cos{A}}\)
\(=\frac{\cos{2A}}{\sin{A}\cos{A}}\) ➜Note \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)

\(=\frac{2\cos{2A}}{2\sin{A}\cos{A}}\) ➜Note লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=\frac{2\cos{2A}}{\sin{2A}}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=2\frac{\cos{2A}}{\sin{2A}}\)
\(=2\cot{2A}\) ➜Note \(\because \frac{\cos{A}}{\sin{A}}=\cot{A}\)

\(\therefore \cot{A}-\tan{A}=2\cot{2A}\)
\(\cot{A}-\tan{A}=2\cot{2A}\)

\(L.S=4\cos^3{x}\sin{3x}+4\sin^3{x}\cos{3x}\)
\(=2\cos^2{x}(2\sin{3x}\cos{x})+2\sin^2{x}(2\cos{3x}\sin{x})\)
\(=2\cos^2{x}\{\sin{(3x+x)}+\sin{(3x-x)}\}+2\sin^2{x}\{\sin{(3x+x)}-\sin{(3x-x)}\}\) ➜Note \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
এবং \(2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\)

\(=2\cos^2{x}\{\sin{4x}+\sin{2x}\}+2\sin^2{x}\{\sin{4x}-\sin{2x}\}\)
\(=2\cos^2{x}\sin{4x}+2\cos^2{x}\sin{2x}+2\sin^2{x}\sin{4x}-2\sin^2{x}\sin{2x}\)
\(=2\sin{4x}(\sin^2{x}+\cos^2{x})+2\sin{2x}(\cos^2{x}-\sin^2{x})\)
\(=2\sin{4x}\times1+2\sin{2x}\cos{2x}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(\cos^2{A}-\sin^2{A}=\cos{2A}\)

\(=2\sin{4x}+\sin{4x}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=3\sin{4x}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\tan{\theta}=\sec{2\alpha}\)
\(L.S=\sin{2\theta}\)
\(=\frac{2\tan{\theta}}{1+\tan^2{\theta}}\) ➜Note \(\because \sin{2A}=\frac{2\tan{A}}{1+\tan^2{A}}\)

\(=\frac{2\sec{2\alpha}}{1+\sec^2{2\alpha}}\) ➜Note \(\because \tan{\theta}=\sec{2\alpha}\)

\(=\frac{\frac{2}{\cos{2\alpha}}}{1+\frac{1}{\cos^2{2\alpha}}}\) ➜Note \(\because \sec{A}=\frac{1}{\cos{A}}\)

\(=\frac{2\cos{2\alpha}}{\cos^2{2\alpha}+1}\) ➜Note লব ও হরকে \(\cos^2{2\alpha}\) দ্বারা গুণ করে।

\(=\frac{2\frac{1-\tan^2{\alpha}}{1+\tan^2{\alpha}}}{\left(\frac{1-\tan^2{\alpha}}{1+\tan^2{\alpha}}\right)^2+1}\) ➜Note \(\because \cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(=\frac{2\frac{1-\tan^2{\alpha}}{1+\tan^2{\alpha}}}{\frac{(1-\tan^2{\alpha})^2}{(1+\tan^2{\alpha})^2}+1}\)
\(=\frac{2(1-\tan^2{\alpha})(1+\tan^2{\alpha})}{(1-\tan^2{\alpha})^2+(1+\tan^2{\alpha})^2}\) ➜Note লব ও হরকে \((1+\tan^2{\alpha})^2\) দ্বারা গুণ করে।

\(=\frac{2(1-\tan^4{\alpha})}{2\times1^2+2\tan^4{\alpha}}\) ➜Note \(\because (a-b)(a+b)=a^2-b^2\)
এবং \((a-b)^2+(a+b)^2=2a^2+2b^2\)

\(=\frac{2(1-\tan^4{\alpha})}{2+2\tan^4{\alpha}}\)
\(=\frac{2(1-\tan^4{\alpha})}{2(1+\tan^4{\alpha})}\)
\(=\frac{1-\tan^4{\alpha}}{1+\tan^4{\alpha}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\tan{x}=\frac{b}{a}\)
\(L.S=a\cos{2x}+b\sin{2x}\)
\(=a\frac{1-\tan^2{x}}{1+\tan^2{x}}+b\frac{2\tan{x}}{1+\tan^2{x}}\) ➜Note \(\because \cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)
এবং \(\sin{2A}=\frac{2\tan{A}}{1+\tan^2{A}}\)

\(=a\frac{1-\left(\frac{b}{a}\right)^2}{1+\left(\frac{b}{a}\right)^2}+b\frac{2\frac{b}{a}}{1+\left(\frac{b}{a}\right)^2}\) ➜Note \(\because \tan{x}=\frac{b}{a}\)

\(=a\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}+b\frac{\frac{2b}{a}}{1+\frac{b^2}{a^2}}\)
\(=a\frac{a^2-b^2}{a^2+b^2}+b\frac{2ab}{a^2+b^2}\) ➜Note উভয় ভগ্নাংশের লব ও হরকে \(a^2\) দ্বারা গুণ করে।

\(=\frac{a^3-ab^2}{a^2+b^2}+\frac{2ab^2}{a^2+b^2}\)
\(=\frac{a^3-ab^2+2ab^2}{a^2+b^2}\)
\(=\frac{a^3+ab^2}{a^2+b^2}\)
\(=\frac{a(a^2+b^2)}{a^2+b^2}\)
\(=a\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\tan{\theta}=\frac{1}{2}\)
\(L.S=10\sin{2\theta}-6\tan{2\theta}+5\cos{2\theta}\)
\(=10\frac{2\tan{\theta}}{1+\tan^2{\theta}}-6\frac{2\tan{\theta}}{1-\tan^2{\theta}}+5\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}\) ➜Note \(\because \sin{2A}=\frac{2\tan{A}}{1+\tan^2{A}}\)
\(\tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)
এবং \(\cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(=10\frac{2\times\frac{1}{2}}{1+\left(\frac{1}{2}\right)^2}-6\frac{2\times\frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}+5\frac{1-\left(\frac{1}{2}\right)^2}{1+\left(\frac{1}{2}\right)^2}\) ➜Note \(\because \tan{\theta}=\frac{1}{2}\)

\(=10\frac{1}{1+\frac{1}{4}}-6\frac{1}{1-\frac{1}{4}}+5\frac{1-\frac{1}{4}}{1+\frac{1}{4}}\)
\(=10\times\frac{4}{4+1}-6\times\frac{4}{4-1}+5\times\frac{4-1}{4+1}\) ➜Note সকল ভগ্নাংশের লব ও হরকে \(4\) দ্বারা গুণ করে।

\(=10\times\frac{4}{5}-6\times\frac{4}{3}+5\times\frac{3}{5}\)
\(=2\times{4}-2\times{4}+3\)
\(=8-8+3\)
\(=3\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\sqrt{3}}{\sin{20^{o}}}-\frac{1}{\cos{20^{o}}}\)
\(=2\times\frac{1}{2}\left(\frac{\sqrt{3}}{\sin{20^{o}}}-\frac{1}{\cos{20^{o}}}\right)\)
\(=2\left(\frac{\frac{\sqrt{3}}{2}}{\sin{20^{o}}}-\frac{\frac{1}{2}}{\cos{20^{o}}}\right)\)
\(=2\left(\frac{\sin{60^{o}}}{\sin{20^{o}}}-\frac{\cos{60^{o}}}{\cos{20^{o}}}\right)\) ➜Note \(\because \frac{\sqrt{3}}{2}=\sin{60^{o}}\)
এবং \(\frac{1}{2}=\cos{60^{o}}\)

\(=2\frac{\sin{60^{o}}\cos{20^{o}}-\cos{60^{o}}\sin{20^{o}}}{\sin{20^{o}}\cos{20^{o}}}\)
\(=2\frac{\sin{(60^{o}-20^{o})}}{\sin{20^{o}}\cos{20^{o}}}\) ➜Note \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)

\(=2\frac{\sin{40^{o}}}{\sin{20^{o}}\cos{20^{o}}}\)
\(=2\frac{\sin{(2\times20^{o})}}{\sin{20^{o}}\cos{20^{o}}}\)
\(=2\frac{2\sin{20^{o}}\cos{20^{o}}}{\sin{20^{o}}\cos{20^{o}}}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=2\times{2}\)
\(=4\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos{4\theta}\)
\(=\cos{2(2\theta)}\)
\(=2\cos^2{2\theta}-1\) ➜Note \(\because \cos{2A}=2\cos^2{A}-1\)

\(=2(\cos{2\theta})^2-1\)
\(=2(2\cos^2{\theta}-1)^2-1\) ➜Note \(\because \cos{2A}=2\cos^2{A}-1\)

\(=2(4\cos^4{\theta}-4\cos^2{\theta}+1)-1\) ➜Note \(\because (a-b)^2=a^2-2ab+b^2\)

\(=8\cos^4{\theta}-8\cos^2{\theta}+2-1\)
\(=8\cos^4{\theta}-8\cos^2{\theta}+1\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sin{8\theta}\)
\(=\sin{2(4\theta)}\)
\(=2\sin{4\theta}\cos{4\theta}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=2\sin{2(2\theta)}\cos{4\theta}\)
\(=2\times2\sin{2\theta}\cos{2\theta}\cos{4\theta}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=4\times2\sin{\theta}\cos{\theta}\cos{2\theta}\cos{4\theta}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=8\sin{\theta}\cos{\theta}\cos{2\theta}\cos{4\theta}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{1}{\sin{10^{o}}}-\frac{\sqrt{3}}{\cos{10^{o}}}\)
\(=2\times\frac{1}{2}\left(\frac{1}{\sin{10^{o}}}-\frac{\sqrt{3}}{\cos{10^{o}}}\right)\)
\(=2\left(\frac{\frac{1}{2}}{\sin{10^{o}}}-\frac{\frac{\sqrt{3}}{2}}{\cos{10^{o}}}\right)\)
\(=2\left(\frac{\sin{30^{o}}}{\sin{10^{o}}}-\frac{\cos{30^{o}}}{\cos{10^{o}}}\right)\) ➜Note \(\because \frac{1}{2}=\sin{30^{o}}\)
এবং \(\frac{\sqrt{3}}{2}=\cos{30^{o}}\)

\(=2\frac{\sin{30^{o}}\cos{10^{o}}-\cos{30^{o}}\sin{10^{o}}}{\sin{10^{o}}\cos{10^{o}}}\)
\(=2\frac{\sin{(30^{o}-10^{o})}}{\sin{10^{o}}\cos{10^{o}}}\) ➜Note \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)

\(=2\frac{\sin{20^{o}}}{\sin{10^{o}}\cos{10^{o}}}\)
\(=2\frac{\sin{(2\times10^{o})}}{\sin{10^{o}}\cos{10^{o}}}\)
\(=2\frac{2\sin{10^{o}}\cos{10^{o}}}{\sin{10^{o}}\cos{10^{o}}}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=2\times{2}\)
\(=4\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sin^2{(60^{o}+A)}+\sin^2{A}+\sin^2{(60^{o}-A)}\)
\(=\frac{1}{2}\{2\sin^2{(60^{o}+A)}+2\sin^2{A}+2\sin^2{(60^{o}-A)}\}\)
\(=\frac{1}{2}\{1-\cos{2(60^{o}+A)}+1-\cos{2A}+1-\cos{2(60^{o}-A)}\}\) ➜Note \(\because 2\sin^2{P}=1-\cos{2P}\)

\(=\frac{1}{2}\{3-\cos{2(60^{o}+A)}-\cos{2(60^{o}-A)}-\cos{2A}\}\)
\(=\frac{3}{2}-\frac{1}{2}\{\cos{(120^{o}-2A)}+\cos{(120^{o}+2A)}\}-\frac{1}{2}\cos{2A}\)
\(=\frac{3}{2}-\frac{1}{2}\times2\cos{120^{o}}\cos{2A}-\frac{1}{2}\cos{2A}\) ➜Note \(\because \cos{(A-B)}+\cos{(A+B)}=2\cos{A}\cos{B}\)

\(=\frac{3}{2}-\cos{120^{o}}\cos{2A}-\frac{1}{2}\cos{2A}\)
\(=\frac{3}{2}+\frac{1}{2}\cos{2A}-\frac{1}{2}\cos{2A}\) ➜Note \(\because \cos{120^{o}}=-\frac{1}{2}\)

\(=\frac{3}{2}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos^3{A}\cos{3A}+\sin^3{A}\sin{3A}\)
\(=\frac{1}{4}\times4\cos^3{A}\cos{3A}+\frac{1}{4}\times4\sin^3{A}\sin{3A}\)
\(=\frac{1}{4}(\cos{3A}+3\cos{A})\cos{3A}+\frac{1}{4}(3\sin{A}-\sin{3A})\sin{3A}\) ➜Note \(\because 4\cos^3{P}=\cos{3P}+3\cos{P}\)
এবং \(4\sin^3{P}=3\sin{P}-\sin{3P}\)

\(=\frac{1}{4}(\cos^2{3A}+3\cos{A}\cos{3A}+3\sin{3A}\sin{A}-\sin^2{3A})\)
\(=\frac{1}{4}\{\cos^2{3A}-\sin^2{3A}+3(\cos{3A}\cos{A}+\sin{3A}\sin{A})\}\)
\(=\frac{1}{4}\{\cos{6A}+3\cos{(3A-A)}\}\) ➜Note \(\because \cos^2{P}-\sin^2{P}=\cos{2P}\)
এবং \(\cos{P}\cos{Q}+\sin{P}\sin{Q}=\cos{(P-Q)}\)

\(=\frac{1}{4}\{\cos{3(2A)}+3\cos{2A}\}\)
\(=\frac{1}{4}\{4\cos^3{2A}-3\cos{2A}+3\cos{2A}\}\) ➜Note \(\because \cos{3P}=4\cos^3{P}-3\cos{P}\)

\(=\frac{1}{4}\{4\cos^3{2A}\}\)
\(=\cos^3{2A}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=16\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}}\)
\(=16\cos{\theta}\cos{2\theta}\cos{4\theta}\cos{7\theta}\) যেখানে, \(\frac{2\pi}{15}=\theta\)
\(=\frac{8}{\sin{\theta}}(2\sin{\theta}\cos{\theta})\cos{2\theta}\cos{4\theta}\cos{7\theta}\)
\(=\frac{8}{\sin{\theta}}\sin{2\theta}\cos{2\theta}\cos{4\theta}\cos{7\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{4}{\sin{\theta}}(2\sin{2\theta}\cos{2\theta})\cos{4\theta}\cos{7\theta}\)
\(=\frac{4}{\sin{\theta}}\sin{4\theta}\cos{4\theta}\cos{7\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{2}{\sin{\theta}}(2\sin{4\theta}\cos{4\theta})\cos{7\theta}\)
\(=\frac{2}{\sin{\theta}}\sin{8\theta}\cos{7\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{1}{\sin{\theta}}(2\sin{8\theta}\cos{7\theta})\)
\(=\frac{1}{\sin{\theta}}\{\sin{(8\theta+7\theta)}+\sin{(8\theta-7\theta)}\}\) ➜Note \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)

\(=\frac{1}{\sin{\theta}}\{\sin{15\theta}+\sin{\theta}\}\)
\(=\frac{1}{\sin{\theta}}\{\sin{2\pi}+\sin{\theta}\}\) ➜Note \(\because \frac{2\pi}{15}=\theta\)
\(\therefore 15\theta=2\pi\)

\(=\frac{1}{\sin{\theta}}\{0+\sin{\theta}\}\) ➜Note \(\because \sin{2\pi}=0\)

\(=\frac{1}{\sin{\theta}}\times\sin{\theta}\)
\(=1\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sin{20^{o}}\sin{40^{o}}\sin{60^{o}}\sin{80^{o}}\)
\(=\sin{20^{o}}\sin{40^{o}}\times\frac{\sqrt{3}}{2}\sin{80^{o}}\) ➜Note \(\because \sin{60^{o}}=\frac{\sqrt{3}}{2}\)

\(=\frac{\sqrt{3}}{2}\sin{20^{o}}\sin{40^{o}}\sin{80^{o}}\)
\(=\frac{\sqrt{3}}{4}(2\sin{80^{o}}\sin{40^{o}})\sin{20^{o}}\)
\(=\frac{\sqrt{3}}{4}\{\cos{(80^{o}-40^{o})}-\cos{(80^{o}+40^{o})}\}\sin{20^{o}}\) ➜Note \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)

\(=\frac{\sqrt{3}}{4}\{\cos{40^{o}}-\cos{120^{o}}\}\sin{20^{o}}\)
\(=\frac{\sqrt{3}}{4}\left\{\cos{40^{o}}+\frac{1}{2}\right\}\sin{20^{o}}\) ➜Note \(\because \cos{120^{o}}=-\frac{1}{2}\)

\(=\frac{\sqrt{3}}{4}\cos{40^{o}}\sin{20^{o}}+\frac{\sqrt{3}}{8}\sin{20^{o}}\)
\(=\frac{\sqrt{3}}{8}(2\cos{40^{o}}\sin{20^{o}})+\frac{\sqrt{3}}{8}\sin{20^{o}}\)
\(=\frac{\sqrt{3}}{8}\{\sin{(40^{o}+20^{o})}-\sin{(40^{o}-20^{o})}\}+\frac{\sqrt{3}}{8}\sin{20^{o}}\) ➜Note \(\because 2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\)

\(=\frac{\sqrt{3}}{8}\{\sin{60^{o}}-\sin{20^{o}}\}+\frac{\sqrt{3}}{8}\sin{20^{o}}\)
\(=\frac{\sqrt{3}}{8}\left\{\frac{\sqrt{3}}{2}-\sin{20^{o}}\right\}+\frac{\sqrt{3}}{8}\sin{20^{o}}\) ➜Note \(\because \sin{60^{o}}=\frac{\sqrt{3}}{2}\)

\(=\frac{3}{16}-\frac{\sqrt{3}}{8}\sin{20^{o}}+\frac{\sqrt{3}}{8}\sin{20^{o}}\)
\(=\frac{3}{16}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\theta=20^{o}\) এবং \(\theta+\alpha+80^{o}=180^{o}\)
\(\Rightarrow 20^{o}+\alpha+80^{o}=180^{o}\)
\(\Rightarrow \alpha+100^{o}=180^{o}\)
\(\Rightarrow \alpha=180^{o}-100^{o}\)
\(\therefore \alpha=80^{o}\)
\(L.S=\tan{\theta}\tan{2\theta}\tan{\alpha}\)
\(=\tan{20^{o}}\tan{40^{o}}\tan{80^{o}}\) ➜Note \(\because \theta=20^{o}\)
এবং \(\alpha=80^{o}\)

\(=\tan{20^{o}}\tan{(60^{o}-20^{o})}\tan{(60^{o}+20^{o})}\) ➜Note \(\because 40^{o}=60^{o}-20^{o}\)
এবং \(80^{o}=60^{o}+20^{o}\)

\(=\tan{20^{o}}\times\frac{\tan{60^{o}}-\tan{20^{o}}}{1+\tan{60^{o}}\tan{20^{o}}}\times\frac{\tan{60^{o}}+\tan{20^{o}}}{1-\tan{60^{o}}\tan{20^{o}}}\) ➜Note \(\because \tan{(A-B)}=\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)
এবং \(\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)

\(=\tan{20^{o}}\times\frac{\sqrt{3}-\tan{20^{o}}}{1+\sqrt{3}\tan{20^{o}}}\times\frac{\sqrt{3}+\tan{20^{o}}}{1-\sqrt{3}\tan{20^{o}}}\) ➜Note \(\because \tan{60^{o}}=\sqrt{3}\)
\(=\tan{20^{o}}\times\frac{(\sqrt{3})^2-\tan^2{20^{o}}}{1^2-(\sqrt{3}\tan{20^{o}})^2}\) ➜Note \(\because (a-b)(a+b)=a^2-b^2\)
\(=\tan{20^{o}}\times\frac{3-\tan^2{20^{o}}}{1-3\tan^2{20^{o}}}\)
\(=\frac{3\tan{20^{o}}-\tan^3{20^{o}}}{1-3\tan^2{20^{o}}}\)
\(=\tan{(3\times20^{o})}\) ➜Note \(\because \frac{3\tan{A}-\tan^3{A}}{1-3\tan^2{A}}=\tan{3A}\)
\(=\tan{60^{o}}\)
\(=\sqrt{3}\) ➜Note \(\because \tan{60^{o}}=\sqrt{3}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(i=\sqrt{-1}\)
\(L.S=(\cos{A}+i\sin{A})^2\)
\(=\cos^2{A}+2\cos{A}i\sin{A}+i^2\sin^2{A}\) ➜Note \(\because (a+b)^2=a^2+2ab+b^2\)

\(=\cos^2{A}+2i\sin{A}\cos{A}-\sin^2{A}\) ➜Note \(\because i=\sqrt{-1}\)
\(\therefore i^2=-1\)

\(=\cos^2{A}-\sin^2{A}+i2\sin{A}\cos{A}\)
\(=\cos{2A}+i\sin{2A}\) ➜Note \(\because \cos^2{P}-\sin^2{P}=\cos{2P}\)
এবং \(2\sin{P}\cos{P}=\sin{2P}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{1-\tan^2{(45^{o}-A)}}{1+\tan^2{(45^{o}-A)}}\)
\(=\cos{\{2(45^{o}-A)\}}\) ➜Note \(\because \frac{1-\tan^2{A}}{1+\tan^2{A}}=\cos{2A}\)

\(=\cos{(90^{o}-2A)}\)
\(=\cos{(90^{o}\times1-2A)}\)
\(=\sin{2A}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ধনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
তাই কোসাইন অনুপাতের পরিবর্তন হয়ে সাইন হয়েছে।

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\sin{x}+\sin{y}=a\) এবং \(\cos{x}+\cos{y}=b\)
ধরি,
\(\sin{x}+\sin{y}=a ........(1)\)
এবং \(\cos{x}+\cos{y}=b ......(2)\)
\((1)\) ও \((2)\) বর্গ করে যোগ করি,
\((\sin{x}+\sin{y})^2+(\cos{x}+\cos{y})^2=a^2+b^2\)
\(\Rightarrow \sin^2{x}+\sin^2{y}+2\sin{x}\sin{y}+\cos^2{x}+\cos^2{y}+2\cos{x}\cos{y}=a^2+b^2\) ➜Note \(\because (a+b)^2=a^2+b^2+2ab\)

\(\Rightarrow (\sin^2{x}+\cos^2{x})+(\sin^2{y}+\cos^2{y})+2(\cos{x}\cos{y}+\sin{x}\sin{y})=a^2+b^2\)
\(\Rightarrow 1+1+2\cos{(\alpha-\beta)}=a^2+b^2\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(\cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)

\(\Rightarrow 2+2\cos{(x-y)}=a^2+b^2\)
\(\Rightarrow 2\{1+\cos{(x-y)}\}=a^2+b^2\)
\(\Rightarrow 2\{2\cos^2{\left(\frac{x-y}{2}\right)}\}=a^2+b^2\) ➜Note \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)

\(\Rightarrow 4\cos^2{\left(\frac{x-y}{2}\right)}=a^2+b^2\)
\(\Rightarrow 4\left\{1-\sin^2{\left(\frac{x-y}{2}\right)}\right\}=a^2+b^2\) ➜Note \(\because \cos^2{A}=1-\sin^2{A}\)

\(\Rightarrow 4-4\sin^2{\left(\frac{x-y}{2}\right)}=a^2+b^2\)
\(\Rightarrow 4-a^2-b^2=4\sin^2{\left(\frac{x-y}{2}\right)}\) ➜Note পক্ষান্তর করে।

\(\Rightarrow 4\sin^2{\left(\frac{x-y}{2}\right)}=4-a^2-b^2\) ➜Note পক্ষান্তর করে।

\(\Rightarrow \sin^2{\left(\frac{x-y}{2}\right)}=\frac{1}{4}(4-a^2-b^2)\)
\(\Rightarrow \sin{\left(\frac{x-y}{2}\right)}=\pm\sqrt{\frac{1}{4}(4-a^2-b^2)}\)
\(\Rightarrow \sin{\frac{1}{2}(x-y)}=\pm\frac{\sqrt{(4-a^2-b^2)}}{2}\)
\(\Rightarrow \frac{1}{\sin{\frac{1}{2}(x-y)}}=\pm\frac{2}{\sqrt{(4-a^2-b^2)}}\) ➜Note ব্যাস্তকরণ করে।

\(\Rightarrow cosec \ {\frac{1}{2}(x-y)}=\pm\frac{2}{\sqrt{(4-a^2-b^2)}}\) ➜Note \(\because \frac{1}{\sin{A}}=cosec \ {A}\)

(প্রমাণিত)

\(L.S=\sqrt{\frac{1-\cos{2\theta}}{1+\cos{2\theta}}}\)
\(=\sqrt{\frac{2\sin^2{\theta}}{2\cos^2{\theta}}}\) ➜Note \(\because 1-\cos{2A}=2\sin^2{A}\)
এবং \(1+\cos{2A}=2\cos^2{A}\)

\(=\sqrt{\frac{\sin^2{\theta}}{\cos^2{\theta}}}\)
\(=\frac{\sin{\theta}}{\cos{\theta}}\)
\(=\tan{\theta}\) ➜Note \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\sin{\theta}+\cos{\theta}}{\sqrt{1+\sin{2\theta}}}\)
\(=\frac{\sin{\theta}+\cos{\theta}}{\sqrt{\sin^2{\theta}+\cos^2{\theta}+\sin{2\theta}}}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)

\(=\frac{\sin{\theta}+\cos{\theta}}{\sqrt{(\sin{\theta}+\cos{\theta})^2}}\) ➜Note \(\because a^2+b^2+2ab=(a+b)^2\)

\(=\frac{\sin{\theta}+\cos{\theta}}{\sin{\theta}+\cos{\theta}}\)
\(=1\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\cos{x}-\cos{2x}}{1-\cos{x}}\)
\(=\frac{\cos{x}-2\cos^2{x}+1}{1-\cos{x}}\) ➜Note \(\because \cos{2A}=2\cos^2{A}-1\)

\(=\frac{2\cos{x}-2\cos^2{x}+1-\cos{x}}{1-\cos{x}}\)
\(=\frac{2\cos{x}(1-\cos{x})+1(1-\cos{x})}{1-\cos{x}}\)
\(=\frac{(1-\cos{x})(2\cos{x}+1)}{1-\cos{x}}\)
\(=2\cos{x}+1\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{1+\cos{2A}+\sin{2A}}{1-\cos{2A}+\sin{2A}}\)
\(=\frac{2\cos^2{A}+2\sin{A}\cos{A}}{2\sin^2{A}+2\sin{A}\cos{A}}\) ➜Note \(\because 1+\cos{2A}=2\cos^2{A}\)
\(\sin{2A}=2\sin{A}\cos{A}\)
এবং \(1-\cos{2A}=2\sin^2{A}\)

\(=\frac{2\cos{A}(\cos{A}+\sin{A})}{2\sin{A}(\cos{A}+\sin{A})}\)
\(=\frac{\cos{A}}{\sin{A}}\)
\(=\cot{A}\) ➜Note \(\because \frac{\cos{A}}{\sin{A}}=\cot{A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=2\cos{2A}+1\)
\(=2(2\cos^2{A}-1)+1\) ➜Note \(\because \cos{2A}=2\cos^2{A}-1\)

\(=4\cos^2{A}-2+1\)
\(=4\cos^2{A}-1\)
\(=(2\cos{A})^2-1^2\)
\(=(2\cos{A}+1)(2\cos{A}-1)\) ➜Note \(\because a^2-b^2=(a+b)(a-b)\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sec{2\theta}-\tan{2\theta}\)
\(=\frac{1}{\cos{2\theta}}-\frac{\sin{2\theta}}{\cos{2\theta}}\) ➜Note \(\because \sec{A}=\frac{1}{\cos{A}}\)
এবং \(\tan{A}=\frac{\sin{A}}{\cos{A}}\)

\(=\frac{1-\sin{2\theta}}{\cos{2\theta}}\)
\(=\frac{\sin^2{\theta}+\cos^2{\theta}-2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)
\(\sin{2A}=2\sin{A}\cos{A}\)
এবং \(\cos{2A}=\cos^2{A}-\sin^2{A}\)

\(=\frac{(\cos{\theta}-\sin{\theta})^2}{(\cos{\theta}-\sin{\theta})(\cos{\theta}+\sin{\theta})}\) ➜Note \(\because a^2+b^2-2ab=(b-a)^2\)
এবং \( a^2-b^2=(a-b)(a+b)\)

\(=\frac{\cos{\theta}-\sin{\theta}}{\cos{\theta}+\sin{\theta}}\)
\(=M.S\)
আবার,
\(M.S=\frac{\cos{\theta}-\sin{\theta}}{\cos{\theta}+\sin{\theta}}\)
\(=\frac{\frac{1}{\sqrt{2}}\cos{\theta}-\frac{1}{\sqrt{2}}\sin{\theta}}{\frac{1}{\sqrt{2}}\cos{\theta}+\frac{1}{\sqrt{2}}\sin{\theta}}\) ➜Note লব ও হরকে \(\frac{1}{\sqrt{2}}\) দ্বারা গুণ করে।

\(=\frac{\cos{\frac{\pi}{4}}\cos{\theta}-\sin{\frac{\pi}{4}}\sin{\theta}}{\cos{\frac{\pi}{4}}\cos{\theta}+\sin{\frac{\pi}{4}}\sin{\theta}}\) ➜Note \(\because \cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}=\frac{1}{\sqrt{2}}\)

\(=\frac{\cos{\left(\frac{\pi}{4}+\theta\right)}}{\cos{\left(\frac{\pi}{4}-\theta\right)}}\) ➜Note \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
এবং \(\cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)

\(=R.S\)
\(\therefore L.S=M.S=R.S\)
(প্রমাণিত)

\(L.S=\tan{A}+\cot{A}\)
\(=\frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}\) ➜Note \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
এবং \(\frac{\cos{A}}{\sin{A}}=\cot{A}\)

\(=\frac{\sin^2{A}+\cos^2{A}}{\sin{A}\cos{A}}\)
\(=\frac{1}{\sin{A}\cos{A}}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)

\(=\frac{2}{2\sin{A}\cos{A}}\) ➜Note লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=2\frac{1}{\sin{2A}}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=2cosec \ {2A}\) ➜Note \(\because \frac{1}{\sin{P}}=cosec \ {P}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\cot{A}-\tan{A}}{\cot{A}+\tan{A}}\)
\(=\frac{\frac{\cos{A}}{\sin{A}}-\frac{\sin{A}}{\cos{A}}}{\frac{\cos{A}}{\sin{A}}+\frac{\sin{A}}{\cos{A}}}\) ➜Note \(\because \frac{\cos{A}}{\sin{A}}=\cot{A}\)
এবং \(\frac{\sin{A}}{\cos{A}}=\tan{A}\)

\(=\frac{\cos^2{A}-\sin^2{A}}{\cos^2{A}+\sin^2{A}}\) ➜Note লব ও হরকে \(\sin{A}\cos{A}\) দ্বারা গুণ করে।

\(=\frac{\cos{2A}}{1}\) ➜Note \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
এবং \(\cos^2{A}+\sin^2{A}=1\)

\(=\cos{2A}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{1-\cos{2\theta}+\sin{2\theta}}{1+\cos{2\theta}+\sin{2\theta}}\)
\(=\frac{2\sin^2{\theta}+2\sin{\theta}\cos{\theta}}{2\cos^2{\theta}+2\sin{\theta}\cos{\theta}}\) ➜Note \(\because 1-\cos{2A}=2\sin^2{A}\)
\(\sin{2A}=2\sin{A}\cos{A}\)
এবং \(1+\cos{2A}=2\cos^2{A}\)

\(=\frac{2\sin{\theta}(\sin{\theta}+\cos{\theta})}{2\cos{\theta}(\cos{\theta}+\sin{\theta})}\)
\(=\frac{\sin{\theta}}{\cos{\theta}}\)
\(=\tan{\theta}\) ➜Note \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\sin{\alpha}-\sqrt{1+\sin{2\alpha}}}{\cos{\alpha}-\sqrt{1+\sin{2\alpha}}}\)
\(=\frac{\sin{\alpha}-\sqrt{\sin^2{\alpha}+\cos^2{\alpha}+2\sin{\alpha}\cos{\alpha}}}{\cos{\alpha}-\sqrt{\sin^2{\alpha}+\cos^2{\alpha}+2\sin{\alpha}\cos{\alpha}}}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(\sin{2A}=2\sin{A}\cos{A}\)

\(=\frac{\sin{\alpha}-\sqrt{(\sin{\alpha}+\cos{\alpha})^2}}{\cos{\alpha}-\sqrt{(\sin{\alpha}+\cos{\alpha})^2}}\) ➜Note \(\because a^2+b^2+2ab=(a+b)^2\)

\(=\frac{\sin{\alpha}-\sin{\alpha}-\cos{\alpha}}{\cos{\alpha}-\sin{\alpha}-\cos{\alpha}}\) ➜Note \(\because \alpha\) ধনাত্মক সূক্ষ্ণকোণ।

\(=\frac{-\cos{\alpha}}{-\sin{\alpha}}\)
\(=\frac{\cos{\alpha}}{\sin{\alpha}}\)
\(=\cot{\alpha}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\cos{(45^{o}+\theta)}}{\cos{(45^{o}-\theta)}}\)
\(=\frac{\cos{45^{o}}\cos{\theta}-\sin{45^{o}}\sin{\theta}}{\cos{45^{o}}\cos{\theta}+\sin{45^{o}}\sin{\theta}}\) ➜Note \(\because \cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
এবং \(\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)

\(=\frac{\frac{1}{\sqrt{2}}\cos{\theta}-\frac{1}{\sqrt{2}}\sin{\theta}}{\frac{1}{\sqrt{2}}\cos{\theta}+\frac{1}{\sqrt{2}}\sin{\theta}}\) ➜Note \(\because \cos{45^{o}}=\sin{45^{o}}=\frac{1}{\sqrt{2}}\)

\(=\frac{\cos{\theta}-\sin{\theta}}{\cos{\theta}+\sin{\theta}}\) ➜Note লব ও হরকে \(\sqrt{2}\) দ্বারা গুণ করে।

\(=\frac{(\cos{\theta}-\sin{\theta})^2}{(\cos{\theta}+\sin{\theta})(\cos{\theta}-\sin{\theta})}\) ➜Note লব ও হরকে \((\cos{\theta}-\sin{\theta})\) দ্বারা গুণ করে।

\(=\frac{\cos^2{\theta}+\sin^2{\theta}-2\cos{\theta}\sin{\theta}}{\cos^2{\theta}-\sin^2{\theta}}\) ➜Note \(\because (a-b)^2=a^2+b^2-2ab\)
এবং \((a+b)(a-b)=a^2-b^2\)

\(=\frac{1-\sin{2\theta}}{\cos{2\theta}}\) ➜Note \(\because \cos^2{A}+\sin^2{A}=1\)
\(2\cos{A}\sin{A}=\sin{2A}\)
এবং \(\cos^2{A}-\sin^2{A}=\cos{2A}\)

\(=\frac{1}{\cos{2\theta}}-\frac{\sin{2\theta}}{\cos{2\theta}}\)
\(=\sec{2\theta}-\tan{2\theta}\) ➜Note \(\because \frac{1}{\cos{A}}=\sec{A}\)
এবং \(\frac{\sin{A}}{\cos{A}}=\tan{A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\tan{2A}\)
\(=\frac{\sin{2A}}{\cos{2A}}\) ➜Note \(\because \tan{P}=\frac{\sin{P}}{\cos{P}}\)

\(=\frac{2\sin{A}\cos{A}}{\cos{2A}}\) ➜Note \(\because \sin{2A}=2\cos{A}\sin{A}\)

\(=\frac{2\frac{\sin{A}}{\cos{A}}\cos^2{A}}{\cos{2A}}\)
\(=\frac{\tan{A}\times2\cos^2{A}}{\cos{2A}}\)
\(=\frac{\tan{A}(1+\cos{2A})}{\cos{2A}}\) ➜Note \(\because 2\cos^2{P}=1+\cos{2P}\)

\(=\frac{1+\cos{2A}}{\cos{2A}}\tan{A}\)
\(=\left\{\frac{1}{\cos{2A}}+\frac{\cos{2A}}{\cos{2A}}\right\}\tan{A}\)
\(=(\sec{2A}+1)\sqrt{\sec^2{A}-1}\) ➜Note \(\because \frac{1}{\cos{P}}=\sec{P}\)
\(\tan{P}=\sqrt{\sec^2{P}-1}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{1+\cos{2\theta}}{\sin{2\theta}}\)
\(=\frac{2\cos^2{\theta}}{2\sin{\theta}\cos{\theta}}\) ➜Note \(\because 1+\cos{2A}=2\cos^2{A}\)
এবং \(\sin{2A}=2\cos{A}\sin{A}\)

\(=\frac{\cos{\theta}}{\sin{\theta}}\)
\(=\cot{\theta}\) ➜Note \(\because \frac{\cos{A}}{\sin{A}}=\cot{A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\tan{\theta}(1+\sec{2\theta})\)
\(=\tan{\theta}(1+\frac{1}{\cos{2\theta}})\) ➜Note \(\because \sec{A}=\frac{1}{\cos{A}}\)

\(=\tan{\theta}\left(1+\frac{1}{\frac{1-\tan^2{A}}{1+\tan^2{A}}}\right)\) ➜Note \(\because \cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(=\tan{\theta}\left(1+\frac{1+\tan^2{A}}{1-\tan^2{A}}\right)\)
\(=\tan{\theta}\left(\frac{1-\tan^2{A}+1+\tan^2{A}}{1-\tan^2{A}}\right)\)
\(=\tan{\theta}\left(\frac{2}{1-\tan^2{A}}\right)\)
\(=\frac{2\tan{\theta}}{1-\tan^2{A}}\)
\(=\tan{2\theta}\) ➜Note \(\because \frac{2\tan{A}}{1-\tan^2{A}}=\tan{2A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\sin{A}+\sin{2A}}{1+\cos{A}+\cos{2A}}\)
\(=\frac{\sin{A}+\sin{2A}}{(1+\cos{2A})+\cos{A}}\)
\(=\frac{\sin{A}+2\sin{A}\cos{A}}{2\cos^2{A}+\cos{A}}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)
এবং \(1+\cos{2A}=2\cos^2{A}\)

\(=\frac{\sin{A}(1+2\cos{A})}{\cos{A}(1+2\cos{A})}\)
\(=\frac{\sin{A}}{\cos{A}}\)
\(=\tan{A}\) ➜Note \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{3\sin{x}-\sin{3x}}{3\cos{x}+\cos{3x}}\)
\(=\frac{3\sin{x}-(3\sin{x}-4sin^3{x})}{3\cos{x}+(4\cos^3{x}-3\cos{x})}\) ➜Note \(\because \sin{3A}=3\sin{A}-4sin^3{A}\)
এবং \(\cos{3A}=4\cos^3{A}-3\cos{A}\)

\(=\frac{3\sin{x}-3\sin{x}+4sin^3{x}}{3\cos{x}+4\cos^3{x}-3\cos{x}}\)
\(=\frac{4sin^3{x}}{4\cos^3{x}}\)
\(=\frac{sin^3{x}}{\cos^3{x}}\)
\(=\tan^3{x}\) ➜Note \(\because \frac{sin{A}}{\cos{A}}=\tan{A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}}-\frac{\cos{\theta}-\sin{\theta}}{\cos{\theta}+\sin{\theta}}\)
\(=\frac{(\cos{\theta}+\sin{\theta})^2-(\cos{\theta}-\sin{\theta})^2}{(\cos{\theta}-\sin{\theta})(\cos{\theta}+\sin{\theta})}\)
\(=\frac{4\cos{\theta}\sin{\theta}}{\cos^2{\theta}-\sin^2{\theta}}\) ➜Note \(\because (a+b)^2-(a-b)^2=4ab\)
এবং \((a-b)(a+b)=a^2-b^2\)

\(=\frac{2\times2\sin{\theta}\cos{\theta}}{\cos{2\theta}}\) ➜Note \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)

\(=\frac{2\sin{2\theta}}{\cos{2\theta}}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=2\tan{2\theta}\) ➜Note \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\cos{A}-\sin{A}}{\cos{A}+\sin{A}}\)
\(=\frac{(\cos{A}-\sin{A})^2}{(\cos{A}+\sin{A})(\cos{A}-\sin{A})}\) ➜Note লব ও হরকে \((\cos{A}-\sin{A})\) দ্বারা গুণ করে।

\(=\frac{\cos^2{A}+\sin^2{A}-2\cos{A}\sin{A}}{\cos^2{A}-\sin^2{A}}\) ➜Note \(\because (a-b)^2=a^2+b^2-2ab\)
এবং \((a+b)(a-b)=a^2-b^2\)

\(=\frac{1-\sin{2A}}{\cos{2A}}\) ➜Note \(\because \cos^2{A}+\sin^2{A}=1\)
\(2\cos{A}\sin{A}=\sin{2A}\)
এবং \(\cos^2{A}-\sin^2{A}=\cos{2A}\)

\(=\frac{1}{\cos{2A}}-\frac{\sin{2A}}{\cos{2A}}\)
\(=\sec{2A}-\tan{2A}\) ➜Note \(\because \frac{1}{\cos{A}}=\sec{A}\)
এবং \(\frac{\sin{A}}{\cos{A}}=\tan{A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sec{x}\)
\(=\frac{1}{\cos{x}}\) ➜Note \(\because \sec{A}=\frac{1}{\cos{A}}\)

\(=\frac{2}{2\cos{x}}\) ➜Note লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=\frac{2}{\sqrt{4\cos^2{x}}}\)
\(=\frac{2}{\sqrt{2\times2\cos^2{x}}}\)
\(=\frac{2}{\sqrt{2(1+\cos{2x})}}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{2}{\sqrt{2+2\cos{2x}}}\)
\(=\frac{2}{\sqrt{2+\sqrt{4\cos^2{2x}}}}\)
\(=\frac{2}{\sqrt{2+\sqrt{2\times2\cos^2{2x}}}}\)
\(=\frac{2}{\sqrt{2+\sqrt{2(1+\cos{4x})}}}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{2}{\sqrt{2+\sqrt{2+2\cos{4x}}}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sec{\frac{3x}{2}}\)
\(=\frac{1}{\cos{\frac{3x}{2}}}\) ➜Note \(\because \sec{A}=\frac{1}{\cos{A}}\)

\(=\frac{2\sqrt{2}}{2\sqrt{2}\cos{\frac{3x}{2}}}\) ➜Note লব ও হরকে \(2\sqrt{2}\) দ্বারা গুণ করে।

\(=\frac{2\sqrt{2}}{\sqrt{8\cos^2{\frac{3x}{2}}}}\)
\(=\frac{2\sqrt{2}}{\sqrt{4\times2\cos^2{\frac{3x}{2}}}}\)
\(=\frac{2\sqrt{2}}{\sqrt{4(1+\cos{3x})}}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{2\sqrt{2}}{\sqrt{4+4\cos{3x})}}\)
\(=\frac{2\sqrt{2}}{\sqrt{4+\sqrt{16\cos^2{3x}}}}\)
\(=\frac{2\sqrt{2}}{\sqrt{4+\sqrt{8\times2\cos^2{3x}}}}\)
\(=\frac{2\sqrt{2}}{\sqrt{4+\sqrt{8(1+\cos{6x})}}}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{2\sqrt{2}}{\sqrt{4+\sqrt{8+8\cos{6x}}}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sec{\frac{5x}{2}}\)
\(=\frac{1}{\cos{\frac{5x}{2}}}\) ➜Note \(\because \sec{A}=\frac{1}{\cos{A}}\)

\(=\frac{2}{2\cos{\frac{5x}{2}}}\) ➜Note লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=\frac{2}{\sqrt{4\cos^2{\frac{5x}{2}}}}\)
\(=\frac{2}{\sqrt{2\times2\cos^2{\frac{5x}{2}}}}\)
\(=\frac{2}{\sqrt{2(1+\cos{5x})}}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{2}{\sqrt{2+2\cos{5x}}}\)
\(=\frac{2}{\sqrt{2+\sqrt{4\cos^2{5x}}}}\)
\(=\frac{2}{\sqrt{2+\sqrt{2\times2\cos^2{5x}}}}\)
\(=\frac{2}{\sqrt{2+\sqrt{2(1+\cos{10x})}}}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{2}{\sqrt{2+\sqrt{2+2\cos{10x}}}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sec{\left(\frac{\pi}{4}+\theta\right)}\sec{\left(\frac{\pi}{4}-\theta\right)}\)
\(=\frac{1}{\cos{\left(\frac{\pi}{4}+\theta\right)}\cos{\left(\frac{\pi}{4}-\theta\right)}}\) ➜Note \(\because \sec{A}=\frac{1}{\cos{A}}\)

\(=\frac{2}{2\cos{\left(\frac{\pi}{4}+\theta\right)}\cos{\left(\frac{\pi}{4}-\theta\right)}}\) ➜Note লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=\frac{2}{\cos{\left(\frac{\pi}{4}+\theta-\frac{\pi}{4}+\theta\right)}+\cos{\left(\frac{\pi}{4}+\theta+\frac{\pi}{4}-\theta\right)}}\) ➜Note \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)

\(=\frac{2}{\cos{2\theta}+\cos{\frac{2\pi}{4}}}\)
\(=\frac{2}{\cos{2\theta}+\cos{\frac{\pi}{2}}}\)
\(=\frac{2}{\cos{2\theta}+0}\) ➜Note \(\because \cos{\frac{\pi}{2}}=0\)

\(=2\frac{1}{\cos{2\theta}}\)
\(=2\sec{2\theta}\) ➜Note \(\because \frac{1}{\cos{A}}=\sec{A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=cosec \ {20^{o}}+\sqrt{3}\sec{20^{o}}\)
\(=\frac{1}{\sin{20^{o}}}+\frac{\sqrt{3}}{\cos{20^{o}}}\) ➜Note \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
এবং \(\sec{A}=\frac{1}{\cos{A}}\)

\(=\frac{\cos{20^{o}}+\sqrt{3}\sin{20^{o}}}{\sin{20^{o}}\cos{20^{o}}}\)
\(=\frac{\frac{1}{2}\cos{20^{o}}+\frac{\sqrt{3}}{2}\sin{20^{o}}}{\frac{1}{2}\sin{20^{o}}\cos{20^{o}}}\) ➜Note লব ও হরকে \(2\) দ্বারা ভাগ করে।

\(=\frac{\sin{30^{o}}\cos{20^{o}}+\cos{30^{o}}\sin{20^{o}}}{\frac{1}{4}(2\sin{20^{o}}\cos{20^{o}})}\) ➜Note \(\because \sin{30^{o}}=\frac{1}{2}\)
এবং \(\cos{30^{o}}=\frac{\sqrt{3}}{2}\)

\(=\frac{\sin{(30^{o}+20^{o})}}{\frac{1}{4}\sin{(2\times20^{o})}}\) ➜Note \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{\sin{50^{o}}}{\frac{1}{4}\sin{40^{o}}}\)
\(=\frac{4\sin{50^{o}}}{\sin{(90^{o}\times1-50^{o})}}\) ➜Note \(\because 40^{o}=90^{o}\times1-50^{o}\)

\(=\frac{4\sin{50^{o}}}{\cos{50^{o}}}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন অনুপাতের পরিবর্তন হয়ে কোসাইন হয়েছে।

\(=4\frac{\sin{50^{o}}}{\cos{50^{o}}}\)
\(=4\tan{50^{o}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sin{\theta}\sin{(60^{o}-\theta)}\sin{(60^{o}+\theta)}\)
\(=\sin{\theta}\{\sin{(60^{o}+\theta)}\sin{(60^{o}-\theta)}\}\)
\(=\sin{\theta}(\sin^2{60^{o}}-\sin^2{\theta})\) ➜Note \(\because \sin{(A+B)}\sin{(A-B)}=\sin^2{A}-\sin^2{B}\)

\(=\sin{\theta}\left\{\left(\frac{\sqrt{3}}{2}\right)^2-\sin^2{\theta}\right\}\) ➜Note \(\because \sin{60^{o}}=\frac{\sqrt{3}}{2}\)

\(=\sin{\theta}\left\{\frac{3}{4}-\sin^2{\theta}\right\}\)
\(=\frac{3}{4}\sin{\theta}-\sin^3{\theta}\)
\(=\frac{1}{4}(3\sin{\theta}-4\sin^3{\theta})\)
\(=\frac{1}{4}\sin{3\theta}\) ➜Note \(\because 3\sin{A}-4\sin^3{A}=\sin{3A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=4\cos{\theta}\cos{\left(\frac{2\pi}{3}+\theta\right)}\cos{\left(\frac{4\pi}{3}+\theta\right)}\)
\(=2\cos{\theta}\{2\cos{\left(\frac{4\pi}{3}+\theta\right)}\cos{\left(\frac{2\pi}{3}+\theta\right)}\}\)
\(=2\cos{\theta}\{\cos{\left(\frac{4\pi}{3}+\theta-\frac{2\pi}{3}-\theta\right)}+\cos{\left(\frac{4\pi}{3}+\theta+\frac{2\pi}{3}+\theta\right)}\}\) ➜Note \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)

\(=2\cos{\theta}\{\cos{\frac{4\pi-2\pi}{3}}+\cos{(\frac{4\pi+2\pi}{3}+2\theta)}\}\)
\(=2\cos{\theta}\{\cos{\frac{2\pi}{3}}+\cos{(\frac{6\pi}{3}+2\theta)}\}\)
\(=2\cos{\theta}\{\cos{\frac{2\pi}{3}}+\cos{(2\pi+2\theta)}\}\)
\(=2\cos{\theta}\{-\frac{1}{2}+\cos{(2\pi+2\theta)}\}\) ➜Note \(\because \cos{\frac{2\pi}{3}}=-\frac{1}{2}\)

\(=-\cos{\theta}+2\cos{(2\pi+2\theta)}\cos{\theta}\)
\(=-\cos{\theta}+\cos{(2\pi+2\theta-\theta)}+\cos{(2\pi+2\theta+\theta)}\) ➜Note \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)

\(=-\cos{\theta}+\cos{(2\pi+\theta)}+\cos{(2\pi+3\theta)}\)
\(=-\cos{\theta}+\cos{\left(\frac{\pi}{2}\times4+\theta\right)}+\cos{\left(\frac{\pi}{2}\times4+3\theta\right)}\) ➜Note \(\because 2\pi+\theta=\frac{\pi}{2}\times4+\theta\)
এবং \(2\pi+3\theta=\frac{\pi}{2}\times4+3\theta\)

\(=-\cos{\theta}+\cos{\theta}+\cos{3\theta}\) ➜Note \(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(4\) একটি জোড় সংখ্যা,
সুতরাং অনুপাতের পরিবর্তন হয়নি।

\(=\cos{3\theta}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos{20^{o}}\cos{40^{o}}\cos{60^{o}}\cos{80^{o}}\)
\(=\cos{20^{o}}\cos{40^{o}}\times\frac{1}{2}\cos{80^{o}}\) ➜Note \(\because \cos{60^{o}}=\frac{1}{2}\)

\(=\frac{1}{2}\cos{20^{o}}\cos{40^{o}}\cos{80^{o}}\)
\(=\frac{1}{4}(2\cos{80^{o}}\cos{40^{o}})\cos{20^{o}}\)
\(=\frac{1}{4}\{\cos{(80^{o}-40^{o})}+\cos{(80^{o}+40^{o})}\}\cos{20^{o}}\) ➜Note \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)

\(=\frac{1}{4}\{\cos{40^{o}}+\cos{120^{o}}\}\cos{20^{o}}\)
\(=\frac{1}{4}\left\{\cos{40^{o}}-\frac{1}{2}\right\}\cos{20^{o}}\) ➜Note \(\because \cos{120^{o}}=-\frac{1}{2}\)

\(=\frac{1}{4}\cos{40^{o}}\cos{20^{o}}-\frac{1}{8}\cos{20^{o}}\)
\(=\frac{1}{8}(2\cos{40^{o}}\cos{20^{o}})-\frac{1}{8}\cos{20^{o}}\)
\(=\frac{1}{8}\{\cos{(40^{o}-20^{o})}+\cos{(40^{o}+20^{o})}\}-\frac{1}{8}\cos{20^{o}}\) ➜Note \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A-B)}\)

\(=\frac{1}{8}\{\cos{20^{o}}+\cos{60^{o}}\}-\frac{1}{8}\cos{20^{o}}\)
\(=\frac{1}{8}\left\{\cos{20^{o}}+\frac{1}{2}\right\}-\frac{1}{8}\cos{20^{o}}\) ➜Note \(\because \cos{60^{o}}=\frac{1}{2}\)

\(=\frac{1}{8}\cos{20^{o}}+\frac{1}{16}-\frac{1}{8}\cos{20^{o}}\)
\(=\frac{3}{16}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=4(\sin^3{10^{o}}+\cos^3{20^{o}})\)
\(=4\sin^3{10^{o}}+4\cos^3{20^{o}}\)
\(=3\sin{10^{o}}-\sin{(3\times10^{o})}+3\cos{20^{o}}+\cos{(3\times20^{o})}\) ➜Note \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
এবং \(4\cos^3{A}=3\cos{A}+\cos{3A}\)

\(=3\sin{10^{o}}-\sin{30^{o}}+3\cos{20^{o}}+\cos{60^{o}}\)
\(=3\sin{10^{o}}-\frac{1}{2}+3\cos{20^{o}}+\frac{1}{2}\) ➜Note \(\because \sin{30^{o}}=\frac{1}{2}\)
এবং \(\cos{60^{o}}=\frac{1}{2}\)

\(=3\sin{10^{o}}+3\cos{20^{o}}\)
\(=3(\sin{10^{o}}+\cos{20^{o}})\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=16\cos{\frac{\pi}{17}}\cos{\frac{2\pi}{17}}\cos{\frac{4\pi}{17}}\cos{\frac{8\pi}{17}}\)
\(=\frac{8\left(2\sin{\frac{\pi}{17}}\cos{\frac{\pi}{17}}\right)\cos{\frac{2\pi}{17}}\cos{\frac{4\pi}{17}}\cos{\frac{8\pi}{17}}}{\sin{\frac{\pi}{17}}}\)
\(=\frac{8\sin{2\frac{\pi}{17}}\cos{\frac{2\pi}{17}}\cos{\frac{4\pi}{17}}\cos{\frac{8\pi}{17}}}{\sin{\frac{\pi}{17}}}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{4\left(2\sin{\frac{2\pi}{17}}\cos{\frac{2\pi}{17}}\right)\cos{\frac{4\pi}{17}}\cos{\frac{8\pi}{17}}}{\sin{\frac{\pi}{17}}}\)
\(=\frac{4\sin{\frac{4\pi}{17}}\cos{\frac{4\pi}{17}}\cos{\frac{8\pi}{17}}}{\sin{\frac{\pi}{17}}}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{2\left(2\sin{\frac{4\pi}{17}}\cos{\frac{4\pi}{17}}\right)\cos{\frac{8\pi}{17}}}{\sin{\frac{\pi}{17}}}\)
\(=\frac{2\sin{\frac{8\pi}{17}}\cos{\frac{8\pi}{17}}}{\sin{\frac{\pi}{17}}}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{\sin{\frac{16\pi}{17}}}{\sin{\frac{\pi}{17}}}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{\sin{\left(\frac{\pi}{2}\times2-\frac{\pi}{17}\right)}}{\sin{\frac{\pi}{17}}}\) ➜Note \(\because \frac{16\pi}{17}=\frac{\pi}{2}\times2-\frac{\pi}{17}\)

\(=\frac{\sin{\frac{\pi}{17}}}{\sin{\frac{\pi}{17}}}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=1\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sin{2x}\tan{2x}\)
\(=\frac{2\tan{x}}{1+\tan^2{x}}\times\frac{2\tan{x}}{1-\tan^2{x}}\) ➜Note \(\because \sin{2A}=\frac{2\tan{A}}{1+\tan^2{A}}\)
এবং \(\tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)

\(=\frac{4\tan^2{x}}{(1+\tan^2{x})(1-\tan^2{x})}\)
\(=\frac{4\tan^2{x}}{1-\tan^4{x}}\) ➜Note \(\because (a+b)(a-b)=a^2-b^2\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=1+\tan{2A}\tan{A}\)
\(=1+\frac{\sin{2A}\sin{A}}{\cos{2A}\cos{A}}\) ➜Note \(\because \tan{P}=\frac{\sin{P}}{\cos{P}}\)

\(=\frac{\cos{2A}\cos{A}+\sin{2A}\sin{A}}{\cos{2A}\cos{A}}\)
\(=\frac{\cos{(2A-A)}}{\cos{2A}\cos{A}}\) ➜Note \(\because \cos{P}\cos{Q}+\sin{P}\sin{Q}=\cos{(P-Q)}\)

\(=\frac{\cos{A}}{\cos{2A}\cos{A}}\)
\(=\frac{1}{\cos{2A}}\)
\(=\sec{2A}\) ➜Note \(\because \frac{1}{\cos{P}}=\sec{P}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sin{(45^{o}+A)}\sin{(45^{o}-A)}\)
\(=\sin^2{45^{o}}-\sin^2{A}\) ➜Note \(\because \sin{(P+Q)}\sin{(P-Q)}=\sin^2{P}-\sin^2{Q}\)

\(=\left(\frac{1}{\sqrt{2}}\right)^2-\sin^2{A}\) ➜Note \(\because \sin{45^{o}}=\frac{1}{\sqrt{2}}\)

\(=\frac{1}{2}-\sin^2{A}\)
\(=\frac{1}{2}(1-2\sin^2{A})\)
\(=\frac{1}{2}\cos{2A}\) ➜Note \(\because 1-2\sin^2{P}=\cos{2P}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\tan{(45^{o}+A)}+\tan{(45^{o}-A)}\)
\(=\frac{\sin{(45^{o}+A)}}{\cos{(45^{o}+A)}}+\frac{\sin{(45^{o}-A)}}{\cos{(45^{o}-A)}}\) ➜Note \(\because \tan{P}=\frac{\sin{P}}{\cos{P}}\)

\(=\frac{\sin{(45^{o}+A)}\cos{(45^{o}-A)}+\cos{(45^{o}+A)}\sin{(45^{o}-A)}}{\cos{(45^{o}+A)}\cos{(45^{o}-A)}}\)
\(=\frac{\sin{(45^{o}+A+45^{o}-A)}}{\cos{(45^{o}+A)}\cos{(45^{o}-A)}}\) ➜Note \(\because \sin{P}\cos{Q}+\cos{P}\sin{Q}=\sin{(P+Q)}\)

\(=\frac{\sin{90^{o}}}{\cos{(45^{o}+A)}\cos{(45^{o}-A)}}\)
\(=\frac{1}{\cos^2{45^{o}}-\sin^2{A}}\) ➜Note \(\because \sin{90^{o}}=1\)
\(\cos{(P+Q)}\cos{P-Q)}=\cos^2{P}-\sin^2{Q}\)

\(=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)^2-\sin^2{A}}\) ➜Note \(\because \cos{45^{o}}=\frac{1}{\sqrt{2}}\)

\(=\frac{1}{\frac{1}{2}-\sin^2{A}}\)
\(=\frac{2}{1-2\sin^2{A}}\) ➜Note লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=2\frac{1}{\cos{2A}}\) ➜Note \(\because 1-2\sin^2{P}=\cos{2P}\)

\(=2\sec{2A}\) ➜Note \(\because \frac{1}{\cos{P}}=\sec{P}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos{A}\sin{(30^{o}+A)}\sin{(30^{o}-A)}\)
\(=\cos{A}(\sin^2{30^{o}}-\sin^2{A})\) ➜Note \(\because \sin{(A+B)}\sin{(A-B)}=\sin^2{A}-\sin^2{B}\)

\(=\cos{A}\left\{\left(\frac{1}{2}\right)^2-\sin^2{A}\right\}\) ➜Note \(\because \sin{30^{o}}=\frac{1}{2}\)

\(=\cos{A}\left\{\frac{1}{4}-\sin^2{A}\right\}\)
\(=\frac{1}{4}\cos{A}-\sin^2{A}\cos{A}\)
\(=\frac{1}{4}\cos{A}-(1-\cos^2{A})\cos{A}\) ➜Note \(\because \sin^2{P}=1-cos^2{P}\)

\(=\frac{1}{4}\cos{A}-\cos{A}+\cos^3{A}\)
\(=\frac{1}{4}(\cos{A}-4\cos{A}+4\cos^3{A})\)
\(=\frac{1}{4}(-3\cos{A}+4\cos^3{A})\)
\(=\frac{1}{4}(4\cos^3{A}-3\cos{A})\)
\(=\frac{1}{4}\cos{3A}\) ➜Note \(\because 4\cos^3{P}-3\cos{P}=cos{3P}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos{\theta}\cos{(60^{o}-\theta)}\cos{(60^{o}+\theta)}\)
\(=\cos{\theta}(\cos^2{60^{o}}-\sin^2{\theta})\) ➜Note \(\because \cos{(A+B)}\cos{(A-B)}=\cos^2{A}-\sin^2{B}\)

\(=\cos{\theta}\left\{\left(\frac{1}{2}\right)^2-\sin^2{\theta}\right\}\) ➜Note \(\because \cos{60^{o}}=\frac{1}{2}\)

\(=\cos{\theta}\left\{\frac{1}{4}-\sin^2{\theta}\right\}\)
\(=\frac{1}{4}\cos{\theta}-\sin^2{\theta}\cos{\theta}\)
\(=\frac{1}{4}\cos{\theta}-(1-\cos^2{\theta})\cos{\theta}\) ➜Note \(\because \sin^2{P}=1-cos^2{P}\)

\(=\frac{1}{4}\cos{\theta}-\cos{\theta}+\cos^3{\theta}\)
\(=\frac{1}{4}(\cos{\theta}-4\cos{\theta}+4\cos^3{\theta})\)
\(=\frac{1}{4}(-\cos{\theta}+4\cos^3{\theta})\)
\(=\frac{1}{4}(4\cos^3{\theta}-3\cos{\theta})\)
\(=\frac{1}{4}\cos{3\theta}\) ➜Note \(\because 4\cos^3{P}-3\cos{P}=cos{3P}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\tan{A}\tan{(60^{o}+A)}\tan{(120^{o}+A)}\)
\(=\tan{A}\tan{(60^{o}+A)}\tan{\{90^{o}\times2-(60^{o}-A)\}}\) ➜Note \(\because 120^{o}+A=90^{o}\times2-(60^{o}-A)\)

\(=-\tan{A}\tan{(60^{o}+A)}\tan{(60^{o}-A)}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং ট্যানজেন্ট অনুপাত ঋনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=-\tan{A}\times\frac{\tan{60^{o}}+\tan{A}}{1-\tan{60^{o}}\tan{A}}\times\frac{\tan{60^{o}}-\tan{A}}{1+\tan{60^{o}}\tan{A}}\) ➜Note \(\because \tan{(P+Q)}=\frac{\tan{P}+\tan{Q}}{1-\tan{P}\tan{Q}}\)
এবং \(\tan{(P-Q)}=\frac{\tan{P}-\tan{Q}}{1+\tan{P}\tan{Q}}\)

\(=-\tan{A}\times\frac{\sqrt{3}+\tan{A}}{1-\sqrt{3}\tan{A}}\times\frac{\sqrt{3}-\tan{A}}{1+\sqrt{3}\tan{A}}\) ➜Note \(\because \tan{60^{o}}=\sqrt{3}\)

\(=-\tan{A}\times\frac{(\sqrt{3}+\tan{A})(\sqrt{3}-\tan{A})}{(1-\sqrt{3}\tan{A})(1+\sqrt{3}\tan{A})}\)
\(=-\tan{A}\times\frac{(\sqrt{3})^2-\tan^2{A}}{1^2-(\sqrt{3}\tan{A})^2}\) ➜Note \(\because (a+b)(a-b)=a^2-b^2\)

\(=-\tan{A}\times\frac{3-\tan^2{A}}{1-3\tan^2{A}}\)
\(=-\frac{3\tan{A}-\tan^3{A}}{1-3\tan^2{A}}\)
\(=-\tan{3A}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=4(\sin^3{25^{o}}+\cos^3{5^{o}})\)
\(=4\sin^3{25^{o}}+4\cos^3{5^{o}}\)
\(=3\sin{25^{o}}-\sin{(3\times25^{o})}+3\cos{5^{o}}+\cos{(3\times5^{o})}\) ➜Note \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
এবং \(4\cos^3{A}=3\cos{A}+\cos{3A}\)

\(=3\sin{25^{o}}-\sin{75^{o}}+3\cos{5^{o}}+\cos{15^{o}}\)
\(=3\sin{25^{o}}-\sin{(90^{o}\times1-15^{o})}+3\cos{(90^{o}\times1-85^{o})}+\cos{15^{o}}\) ➜Note \(\because 5^{o}=90^{o}\times1-85^{o}\)
এবং \(75^{o}=90^{o}\times1-15^{o}\)

\(=3\sin{25^{o}}-\cos{15^{o}}+3\sin{85^{o}}+\cos{15^{o}}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং সাইন এবং কোসাইন উভয় অনুপাত ধনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন এবং কোসাইন অনুপাতের পরিবর্তন হয়ে যথাক্রমে কোসাইন এবং সাইন হয়েছে।

\(=3(\sin{25^{o}}+\sin{85^{o}})\)
\(=3\times2\sin{\frac{85^{o}+25^{o}}{2}}\cos{\frac{85^{o}-25^{o}}{2}}\) ➜Note \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)

\(=6\sin{\frac{110^{o}}{2}}\cos{\frac{60^{o}}{2}}\)
\(=6\sin{55^{o}}\cos{30^{o}}\)
\(=6\sin{55^{o}}\times\frac{\sqrt{3}}{2}\) ➜Note \(\because \cos{30^{o}}=\frac{\sqrt{3}}{2}\)

\(=3\sqrt{3}\sin{55^{o}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\cos^3{x}+\sin^3{x}}{\cos{x}+\sin{x}}\)
\(=\frac{(\cos{x}+\sin{x})(\cos^2{x}-\cos{x}\sin{x}+\sin^2{x})}{\cos{x}+\sin{x}}\) ➜Note \(\because a^3+b^3=(a+b)(a^2-ab+b^2)\)

\(=\cos^2{x}-\cos{x}\sin{x}+\sin^2{x}\)
\(=\sin^2{x}+\cos^2{x}-\cos{x}\sin{x}\)
\(=1-\cos{x}\sin{x}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)

\(=1-\frac{1}{2}(2\sin{x}\cos{x})\)
\(=1-\frac{1}{2}\sin{2x}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\tan^2{\left(\theta+\frac{\pi}{4}\right)}-1}{\tan^2{\left(\theta+\frac{\pi}{4}\right)}+1}\)
\(=-\frac{1-\tan^2{\left(\theta+\frac{\pi}{4}\right)}}{1+\tan^2{\left(\theta+\frac{\pi}{4}\right)}}\)
\(=-\cos{\left\{2\left(\theta+\frac{\pi}{4}\right)\right\}}\) ➜Note \(\because \frac{1-\tan^2{A}}{1+\tan^2{A}}=\cos{2A}\)

\(=-\cos{\left(2\theta+\frac{\pi}{2}\right)}\)
\(=-\cos{\left(\frac{\pi}{2}\times1+2\theta\right)}\)
\(=\sin{2\theta}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ঋনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
তাই কোসাইন অনুপাতের পরিবর্তন হয়ে সাইন হয়েছে।

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\tan{\left(\alpha+\frac{\pi}{3}\right)}+\tan{\left(\alpha-\frac{\pi}{3}\right)}\)
\(=\frac{\sin{\left(\alpha+\frac{\pi}{3}\right)}}{\cos{\left(\alpha+\frac{\pi}{3}\right)}}+\frac{\sin{\left(\alpha-\frac{\pi}{3}\right)}}{\cos{\left(\alpha-\frac{\pi}{3}\right)}}\) ➜Note \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)

\(=\frac{\sin{\left(\alpha+\frac{\pi}{3}\right)}\cos{\left(\alpha-\frac{\pi}{3}\right)}+\cos{\left(\alpha+\frac{\pi}{3}\right)}\sin{\left(\alpha-\frac{\pi}{3}\right)}}{\cos{\left(\alpha+\frac{\pi}{3}\right)}\cos{\left(\alpha-\frac{\pi}{3}\right)}}\)
\(=\frac{\sin{\left(\alpha+\frac{\pi}{3}+\alpha-\frac{\pi}{3}\right)}}{\cos{\left(\alpha+\frac{\pi}{3}\right)}\cos{\left(\alpha-\frac{\pi}{3}\right)}}\) ➜Note \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)

\(=\frac{\sin{2\alpha}}{\cos^2{\frac{\pi}{3}}-\sin^2{\alpha}}\) ➜Note \(\because \cos{(A+B)}\cos{(A-B)}=\cos^2{B}-\sin^2{A}\)

\(=\frac{\sin{2\alpha}}{\left(\frac{1}{2}\right)^2-\sin^2{\alpha}}\) ➜Note \(\because \cos{\frac{\pi}{3}}=\frac{1}{2}\)

\(=\frac{\sin{2\alpha}}{\frac{1}{4}-\sin^2{\alpha}}\)
\(=\frac{4\sin{2\alpha}}{1-4\sin^2{\alpha}}\) ➜Note লব ও হরকে \(4\) দ্বারা গুণ করে।

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=4(\cos^3{10^{o}}+\sin^3{20^{o}})\)
\(=4\cos^3{10^{o}}+4\sin^3{20^{o}}\)
\(=3\cos{10^{o}}+\cos{(3\times10^{o})}+3\sin{20^{o}}-\sin{(3\times20^{o})}\) ➜Note \(\because 4\cos^3{A}=3\cos{A}+\cos{3A}\)
এবং \(4\sin^3{A}=3\sin{A}-\sin{3A}\)

\(=3\cos{10^{o}}+\cos{30^{o}}+3\sin{20^{o}}-\sin{60^{o}}\)
\(=3\cos{10^{o}}+\frac{\sqrt{3}}{2}+3\sin{20^{o}}-\frac{\sqrt{3}}{2}\) ➜Note \(\because \cos{30^{o}}=\frac{\sqrt{3}}{2}\)
এবং \(\sin{60^{o}}=\frac{\sqrt{3}}{2}\)

\(=3\cos{10^{o}}+3\sin{20^{o}}\)
\(=3(\cos{10^{o}}+\sin{20^{o}})\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos{nA}\cos{(n+2)A}-\cos^2{(n+1)A}+\sin^2{A}\)
\(=\cos{nA}\cos{(n+2)A}-\frac{1}{2}\{2\cos^2{(n+1)A}-2\sin^2{A}\}\)
\(=\cos{nA}\cos{(n+2)A}-\frac{1}{2}\{1+\cos{2(n+1)A}-1+\cos{2A}\}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)
এবং \(2\sin^2{A}=1-\cos{2A}\)

\(=\cos{nA}\cos{(n+2)A}-\frac{1}{2}\{\cos{(2n+2)A}+\cos{2A}\}\)
\(=\cos{nA}\cos{(n+2)A}-\frac{1}{2}\{2\cos{\frac{(2n+2)A+2A}{2}}\cos{\frac{(2n+2)A-2A}{2}}\}\) ➜Note \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)

\(=\cos{nA}\cos{(n+2)A}-\cos{\frac{2nA+2A+2A}{2}}\cos{\frac{2nA+2A-2A}{2}}\)
\(=\cos{nA}\cos{(n+2)A}-\cos{\frac{2nA+4A}{2}}\cos{\frac{2nA}{2}}\)
\(=\cos{nA}\cos{(n+2)A}-\cos{\frac{2(n+2)A}{2}}\cos{nA}\)
\(=\cos{nA}\cos{(n+2)A}-\cos{nA}\cos{(n+2)A}\)
\(=0\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sin^2{\left(\frac{\pi}{8}+\frac{\theta}{2}\right)}-\sin^2{\left(\frac{\pi}{8}-\frac{\theta}{2}\right)}\)
\(=\sin{\left(\frac{\pi}{8}+\frac{\theta}{2}+\frac{\pi}{8}-\frac{\theta}{2}\right)}\sin{\left(\frac{\pi}{8}+\frac{\theta}{2}-\frac{\pi}{8}+\frac{\theta}{2}\right)}\) ➜Note \(\because \sin^2{A}-\sin^2{B}=\sin{(A+B)}\sin{(A-B)}\)

\(=\sin{\left(\frac{2\pi}{8}\right)}\sin{\left(\frac{2\theta}{2}\right)}\)
\(=\sin{\left(\frac{\pi}{4}\right)}\sin{\theta}\)
\(=\frac{1}{\sqrt{2}}\sin{\theta}\) ➜Note \(\because \sin{\left(\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos^6{\theta}+\sin^6{\theta}\)
\(=(\sin^2{\theta})^3+(\cos^2{\theta})^3\)
\(=(\sin^2{\theta}+\cos^2{\theta})^3-3\sin^2{\theta}\cos^2{\theta}(\sin^2{\theta}+\cos^2{\theta})\) ➜Note \(\because a^3+b^3=(a+b)^3-3ab(a+b)\)

\(=1^3-3\sin^2{\theta}\cos^2{\theta}.1\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)

\(=1-3\sin^2{\theta}\cos^2{\theta}\)
\(=1-\frac{3}{4}(2\sin{\theta}\cos{\theta})^2\)
\(=1-\frac{3}{4}(\sin{2\theta})^2\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=1-\frac{3}{4}\sin^2{2\theta}\)
\(=M.S\)
আবার,
\(M.S=1-\frac{3}{4}\sin^2{2\theta}\)
\(=1-\frac{3}{4}(1-\cos^2{2\theta})\) ➜Note \(\because \sin^2{A}=1-\cos^2{A}\)

\(=1-\frac{3}{4}+\frac{3}{4}\cos^2{2\theta}\)
\(=\frac{4-3}{4}+\frac{3}{4}\cos^2{2\theta}\)
\(=\frac{1}{4}+\frac{3}{4}\cos^2{2\theta}\)
\(=\frac{1}{4}(1+3\cos^2{2\theta})\)
\(=R.S\)
\(\therefore L.S=M.S=R.S\)
(প্রমাণিত)

\(L.S=\sin{5\theta}\)
\(=\sin{(3\theta+2\theta)}\)
\(=\sin{3\theta}\cos{2\theta}+\cos{3\theta}\sin{2\theta}\) ➜Note \(\because \sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)

\(=(3\sin{\theta}-4\sin^3{\theta})(1-2\sin^2{\theta})+(4\cos^3{\theta}-3\cos{\theta})2\sin{\theta}\cos{\theta}\) ➜Note \(\because \sin{3A}=3\sin{A}-4\sin^3{A}\)
\(\cos{2A}=1-2\sin^2{A}\)
\(\cos{3A}=4\cos^3{A}-3\cos{A}\)
এবং \(\sin{2A}=2\sin{A}\cos{A}\)

\(=3\sin{\theta}-4\sin^3{\theta}-6\sin^3{\theta}+8\sin^5{\theta}+2\sin{\theta}\cos^2{\theta}(4\cos^2{\theta}-3)\)
\(=8\sin^5{\theta}-10\sin^3{\theta}+3\sin{\theta}+2\sin{\theta}(1-\sin^2{\theta})\{4(1-\sin^2{\theta})-3\}\) ➜Note \(\because \cos^2{A}=1-\sin^2{A}\)

\(=8\sin^5{\theta}-10\sin^3{\theta}+3\sin{\theta}+2\sin{\theta}(1-\sin^2{\theta})\{4-4\sin^2{\theta}-3\}\)
\(=8\sin^5{\theta}-10\sin^3{\theta}+3\sin{\theta}+2\sin{\theta}(1-\sin^2{\theta})(1-4\sin^2{\theta})\)
\(=8\sin^5{\theta}-10\sin^3{\theta}+3\sin{\theta}+2\sin{\theta}(1-\sin^2{\theta}-4\sin^2{\theta}+4\sin^4{\theta})\)
\(=8\sin^5{\theta}-10\sin^3{\theta}+3\sin{\theta}+2\sin{\theta}(1-5\sin^2{\theta}+4\sin^4{\theta})\)
\(=8\sin^5{\theta}-10\sin^3{\theta}+3\sin{\theta}+2\sin{\theta}-10\sin^3{\theta}+8\sin^5{\theta}\)
\(=16\sin^5{\theta}-20\sin^3{\theta}+5\sin{\theta}\)
\(=R.S\)
\(\therefore L.S=M.S=R.S\)
(প্রমাণিত)

\(L.S=\cos{5\theta}\)
\(=\cos{(3\theta+2\theta)}\)
\(=\cos{3\theta}\cos{2\theta}-\sin{3\theta}\sin{2\theta}\) ➜Note \(\because \cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)

\(=(4\cos^3{\theta}-3\cos{\theta})(2\cos^2{\theta}-1)-(3\sin{\theta}-4\sin^3{\theta})2\sin{\theta}\cos{\theta}\) ➜Note \(\because \cos{3A}=4\cos^3{A}-3\cos{A}\)
\(\cos{2A}=2\cos^2{A}-1\)
\(\sin{3A}=3\sin{A}-4\sin^3{A}\)
এবং \(\sin{2A}=2\sin{A}\cos{A}\)

\(=8\cos^5{\theta}-6\cos^3{\theta}-4\cos^3{\theta}+3\cos{\theta}-2\sin^2{\theta}\cos{\theta}(3-4\sin^2{\theta})\)
\(=8\cos^5{\theta}-10\cos^3{\theta}+3\cos{\theta}-2\cos{\theta}(1-\cos^2{\theta})\{3-4(1-\cos^2{\theta})\}\) ➜Note \(\because \sin^2{A}=1-\cos^2{A}\)

\(=8\cos^5{\theta}-10\cos^3{\theta}+3\cos{\theta}-2\cos{\theta}(1-\cos^2{\theta})\{3-4+4\cos^2{\theta}\}\)
\(=8\cos^5{\theta}-10\cos^3{\theta}+3\cos{\theta}-2\cos{\theta}(1-\cos^2{\theta})(4\cos^2{\theta}-1)\)
\(=8\cos^5{\theta}-10\cos^3{\theta}+3\cos{\theta}-2\cos{\theta}(4\cos^2{\theta}-4\cos^4{\theta}-1+\cos^2{\theta})\)
\(=8\cos^5{\theta}-10\cos^3{\theta}+3\cos{\theta}-2\cos{\theta}(5\cos^2{\theta}-4\cos^4{\theta}-1)\)
\(=8\cos^5{\theta}-10\cos^3{\theta}+3\cos{\theta}-10\cos^3{\theta}+8\cos^5{\theta}+2\cos{\theta}\)
\(=16\cos^5{\theta}-20\cos^3{\theta}+5\cos{\theta}\)
\(=R.S\)
\(\therefore L.S=M.S=R.S\)
(প্রমাণিত)

\(L.S=\cos^2{(A-120^{o})}+\cos^2{A}+\cos^2{(A+120^{o})}\)
\(=\frac{1}{2}\{2\cos^2{(A-120^{o})}+2\cos^2{A}+2\cos^2{(A+120^{o})}\}\)
\(=\frac{1}{2}\{1+\cos{2(A-120^{o})}+1+\cos{2A}+1+\cos{2(A+120^{o})}\}\) ➜Note \(\because 2\cos^2{P}=1+\cos{2P}\)

\(=\frac{1}{2}\{3+\cos{2(A-120^{o})}+\cos{2(A+120^{o})}+\cos{2A}\}\)
\(=\frac{3}{2}+\frac{1}{2}\{\cos{(2A-240^{o})}+\cos{(2A+240^{o})}\}+\frac{1}{2}\cos{2A}\)
\(=\frac{3}{2}+\frac{1}{2}\times2\cos{2A}\cos{240^{o}}+\frac{1}{2}\cos{2A}\) ➜Note \(\because \cos{(A-B)}+\cos{(A+B)}=2\cos{A}\cos{B}\)

\(=\frac{3}{2}+\cos{2A}\cos{(90^{o}\times2+60^{o})}+\frac{1}{2}\cos{2A}\) ➜Note \(\because 240^{o}=90^{o}\times2+60^{o}\)

\(=\frac{3}{2}-\cos{2A}\cos{60^{o}}+\frac{1}{2}\cos{2A}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি তৃতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ঋনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=\frac{3}{2}-\frac{1}{2}\cos{2A}+\frac{1}{2}\cos{2A}\) ➜Note \(\because \cos{60^{o}}=\frac{1}{2}\)

\(=\frac{3}{2}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sin^3{x}+\sin^3{(120^{o}+x)}+\sin^3{(240^{o}+x)}\)
\(=\frac{1}{4}\{4\sin^3{x}+4\sin^3{(120^{o}+x)}+4\sin^3{(240^{o}+x)}\}\)
\(=\frac{1}{4}\{3\sin{x}-\sin{3x}+3\sin{(120^{o}+x)}-\sin{3(120^{o}+x)}+3\sin{(240^{o}+x)}-\sin{3(240^{o}+x)}\}\) ➜Note \(\because 4\sin^3{P}=3\sin{P}-\sin{3P}\)

\(=\frac{1}{4}[3\sin{x}-\sin{3x}+3\{\sin{(240^{o}+x)}+\sin{(120^{o}+x)}\}-\sin{(360^{o}+3x)}-\sin{(720^{o}+3x)}]\)
\(=\frac{1}{4}[3\sin{x}-\sin{3x}+3\{2\sin{\frac{240^{o}+x+120^{o}+x}{2}}\cos{\frac{240^{o}+x-120^{o}-x}{2}}\}-\sin{(90^{o}\times4+3x)}-\sin{(90^{o}\times8+3x)}]\)➜Note \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)

\(=\frac{1}{4}[3\sin{x}-\sin{3x}+6\sin{\frac{360^{o}+2x}{2}}\cos{\frac{120^{o}}{2}}-\sin{3x}-\sin{3x}]\) ➜Note \(\because\) উভয় ক্ষেত্রে কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(4\) এবং \(8\) উভয়েই জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=\frac{1}{4}[3\sin{x}+6\sin{\frac{2(180^{o}+x)}{2}}\cos{60^{o}}-3\sin{3x}]\)
\(=\frac{1}{4}[3\sin{x}+6\sin{(180^{o}+x)}\times\frac{1}{2}-3\sin{3x}]\) ➜Note \(\because \cos{60^{o}}=\frac{1}{2}\)

\(=\frac{1}{4}[3\sin{x}+3\sin{(90^{o}\times2+x)}-3\sin{3x}]\)
\(=\frac{1}{4}[3\sin{x}-3\sin{x}-3\sin{3x}]\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি তৃতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ঋনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=\frac{1}{4}[-3\sin{3x}]\)
\(=-\frac{3}{4}\sin{3x}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos^3{x}+\cos^3{(60^{o}-x)}+\cos^3{(60^{o}+x)}\)
\(=\frac{1}{4}\{4\cos^3{x}+4\cos^3{(60^{o}-x)}+4\cos^3{(60^{o}+x)}\}\)
\(=\frac{1}{4}\{\cos{3x}+3\cos{x}+\cos{3(60^{o}-x)}+3\cos{(60^{o}-x)}+\cos{3(60^{o}+x)}+3\cos{(60^{o}+x)}\}\) ➜Note \(\because 4\cos^3{P}=\cos{3P}+3\cos{3P}\)

\(=\frac{1}{4}[\cos{3x}+3\cos{x}+\cos{(180^{o}-3x)}+\cos{(180^{o}+3x)}+3\{\cos{(60^{o}-x)}+\cos{(60^{o}+x)}\}]\)
\(=\frac{1}{4}[\cos{3x}+3\cos{x}+\cos{(90^{o}\times2-3x)}+\cos{(90^{o}\times2+3x)}+3\{2\cos{\frac{60^{o}+x+60^{o}-x}{2}}\cos{\frac{60^{o}+x-60^{o}+x}{2}}\}]\)➜Note \(\because \cos{D}+\cos{C}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)

\(=\frac{1}{4}[\cos{3x}+3\cos{x}-\cos{3x}-\cos{3x}+6\cos{\frac{120^{o}}{2}}\cos{\frac{2x}{2}}]\) ➜Note \(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় এবং তৃতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ঋনাত্মক।
আবার, \(90^{o}\) এর সহগুণক উভয় ক্ষেত্রে \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=\frac{1}{4}[3\cos{x}-\cos{3x}+6\cos{60^{o}}\cos{x}]\)
\(=\frac{1}{4}[3\cos{x}-\cos{3x}+6\times\frac{1}{2}\cos{x}]\) ➜Note \(\because \cos{60^{o}}=\frac{1}{2}\)

\(=\frac{1}{4}[3\cos{x}-\cos{3x}+3\cos{x}]\)
\(=\frac{1}{4}(6\cos{x}-\cos{3x})\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos^3{x}+\cos^3{(120^{o}+x)}+\cos^3{(240^{o}+x)}\)
\(=\frac{1}{4}\{4\cos^3{x}+4\cos^3{(120^{o}+x)}+4\cos^3{(240^{o}+x)}\}\)
\(=\frac{1}{4}\{\cos{3x}+3\cos{x}+\cos{3(120^{o}+x)}+3\cos{(120^{o}+x)}+\cos{3(240^{o}+x)}+3\cos{(240^{o}+x)}\}\) ➜Note \(\because 4\cos^3{P}=\cos{3P}+3\cos{3P}\)

\(=\frac{1}{4}[\cos{3x}+3\cos{x}+\cos{(360^{o}+3x)}+\cos{(720^{o}+3x)}+3\{\cos{(120^{o}+x)}+\cos{(240^{o}+x)}\}]\)
\(=\frac{1}{4}[\cos{3x}+3\cos{x}+\cos{(90^{o}\times4+3x)}+\cos{(90^{o}\times8+3x)}+3\{2\cos{\frac{240^{o}+x+120^{o}+x}{2}}\cos{\frac{240^{o}+x-120^{o}-x}{2}}\}]\)➜Note \(\because \cos{D}+\cos{C}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)

\(=\frac{1}{4}[\cos{3x}+3\cos{x}+\cos{3x}+\cos{3x}+6\cos{\frac{360^{o}+2x}{2}}\cos{\frac{120^{o}}{2}}]\) ➜Note \(\because\) উভয় ক্ষেত্রে কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ধনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(4\) এবং \(8\) উভয়েই জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=\frac{1}{4}[3\cos{3x}+3\cos{x}+6\cos{\frac{2(180^{o}+x)}{2}}\cos{60^{o}}]\)
\(=\frac{1}{4}[3\cos{3x}+3\cos{x}+6\cos{(180^{o}+x)}\times\frac{1}{2}]\) ➜Note \(\because \cos{60^{o}}=\frac{1}{2}\)

\(=\frac{1}{4}[3\cos{3x}+3\cos{x}+3\cos{(90^{o}\times2+x)}]\)
\(=\frac{1}{4}[3\cos{3x}+3\cos{x}-3\cos{x}]\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি তৃতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ঋনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=\frac{1}{4}[3\cos{3x}]\)
\(=\frac{3}{4}\cos{3x}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sin^4{x}+\cos^4{x}\)
\(=(\sin^2{x})^2+(\cos^2{x})^2\)
\(=(\sin^2{x}+\cos^2{x})^2-2\sin^2{x}\cos^2{x}\) ➜Note \(\because a^2+b^2=(a+b)^2-2ab\)

\(=(1)^2-\frac{1}{2}(2\sin{x}\cos{x})^2\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)

\(=1-\frac{1}{2}(\sin{2x})^2\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=1-\frac{1}{2}\sin^2{2x}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos^4{\theta}-\sin^4{\theta}\)
\(=(\cos^2{\theta})^2-(\sin^2{\theta})^2\)
\(=(\cos^2{\theta}-\sin^2{\theta})(\cos^2{\theta}+\sin^2{\theta})\) ➜Note \(\because a^2-b^2=(a-b)(a+b)\)

\(=\cos{2\theta}.1\) ➜Note \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
এবং \(\cos^2{A}+\sin^2{A}=1\)

\(=\cos{2\theta}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos^4{x}\)
\(=\frac{1}{4}(2\cos^2{x})^2\)
\(=\frac{1}{4}(1+\cos{2x})^2\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{1}{4}(1^2+\cos^2{2x}+2\cos{2x})\) ➜Note \(\because (a+b)^2=a^2+b^2+2ab\)

\(=\frac{1}{4}(1+\cos^2{2x}+2\cos{2x})\)
\(=\frac{1}{4}+\frac{1}{4}\cos^2{2x}+\frac{1}{2}\cos{2x}\)
\(=\frac{1}{4}+\frac{1}{2}\cos{2x}+\frac{1}{4}\cos^2{2x}\)
\(=\frac{1}{4}+\frac{1}{2}\cos{2x}+\frac{1}{8}(2\cos^2{2x})\)
\(=\frac{1}{4}+\frac{1}{2}\cos{2x}+\frac{1}{8}(1+\cos{4x})\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{1}{4}+\frac{1}{2}\cos{2x}+\frac{1}{8}+\frac{1}{8}\cos{4x}\)
\(=\frac{1}{4}+\frac{1}{8}+\frac{1}{2}\cos{2x}+\frac{1}{8}\cos{4x}\)
\(=\frac{2+1}{8}+\frac{1}{2}\cos{2x}+\frac{1}{8}\cos{4x}\)
\(=\frac{3}{8}+\frac{1}{2}\cos{2x}+\frac{1}{8}\cos{4x}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos^3{A}\cos{3A}+\sin^3{A}\sin{3A}\)
\(=\frac{1}{4}\times4\cos^3{A}\cos{3A}+\frac{1}{4}\times4\sin^3{A}\sin{3A}\)
\(=\frac{1}{4}(\cos{3A}+3\cos{A})\cos{3A}+\frac{1}{4}(3\sin{A}-\sin{3A})\sin{3A}\) ➜Note \(\because 4\cos^3{P}=\cos{3P}+3\cos{P}\)
এবং \(4\sin^3{P}=3\sin{P}-\sin{3P}\)

\(=\frac{1}{4}(\cos^2{3A}+3\cos{A}\cos{3A}+3\sin{3A}\sin{A}-\sin^2{3A})\)
\(=\frac{1}{4}\{\cos^2{3A}-\sin^2{3A}+3(\cos{3A}\cos{A}+\sin{3A}\sin{A})\}\)
\(=\frac{1}{4}\{\cos{6A}+3\cos{(3A-A)}\}\) ➜Note \(\because \cos^2{P}-\sin^2{P}=\cos{2P}\)
এবং \(\cos{P}\cos{Q}+\sin{P}\sin{Q}=\cos{(P-Q)}\)

\(=\frac{1}{4}\{\cos{3(2A)}+3\cos{2A}\}\)
\(=\frac{1}{4}\{4\cos^3{2A}-3\cos{2A}+3\cos{2A}\}\) ➜Note \(\because \cos{3P}=4\cos^3{P}-3\cos{P}\)

\(=\frac{1}{4}\{4\cos^3{2A}\}\)
\(=\cos^3{2A}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=4\cos^3{x}\sin{3x}+4\sin^3{x}\cos{3x}\)
\(=(\cos{3x}+3\cos{x})\sin{3x}+(3\sin{x}-\sin{3x})\cos{3x}\) ➜Note \(\because 4\cos^3{P}=\cos{3P}+3\cos{P}\)
এবং \(4\sin^3{P}=3\sin{P}-\sin{3P}\)

\(=\sin{3x}\cos{3x}+3\sin{3x}\cos{x}+3\cos{3x}\sin{x}-\sin{3x}\cos{3x}\)
\(=3(\sin{3x}\cos{x}+\cos{3x}\sin{x})\)
\(=3\sin{(3x+x)}\) ➜Note \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)

\(=3\sin{4x}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\cos^4{\frac{\pi}{8}}+\cos^4{\frac{3\pi}{8}}+\cos^4{\frac{5\pi}{8}}+\cos^4{\frac{7\pi}{8}}\)
\(=\cos^4{\frac{\pi}{8}}+\cos^4{\left(\frac{\pi}{2}\times1-\frac{\pi}{8}\right)}+\cos^4{\left(\frac{\pi}{2}\times1+\frac{\pi}{8}\right)}+\cos^4{\left(\frac{\pi}{2}\times2-\frac{\pi}{8}\right)}\) ➜Note \(\because \frac{3\pi}{8}=\frac{\pi}{2}\times1-\frac{\pi}{8}\)
\(\frac{5\pi}{8}=\frac{\pi}{2}\times1+\frac{\pi}{8}\)
এবং \(\frac{7\pi}{8}=\frac{\pi}{2}\times2-\frac{\pi}{8}\)

\(=\cos^4{\frac{\pi}{8}}+\sin^4{\frac{\pi}{8}}+(-1)^4\sin^4{\frac{\pi}{8}}+(-1)^4\cos^4{\frac{\pi}{8}}\) ➜Note \(\because\) দ্বিতীয় পদে কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে এবং তৃতীয় ও চতুর্থ উভয় পদে ইহা দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং দ্বিতীয় পদে কোসাইন অনুপাত ধনাত্মক এবং তৃতীয় ও চতুর্থ উভয় পদে কোসাইন অনুপাত ঋনাত্মক।
আবার, দ্বিতীয় ও তৃতীয় পদে \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
তাই কোসাইন অনুপাতের পরিবর্তন হয়ে সাইন হয়েছে
এবং চতুর্থ পদে \(\frac{\pi}{2}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=2\cos^4{\frac{\pi}{8}}+2\sin^4{\frac{\pi}{8}}\)
\(=2\left\{\sin^4{\frac{\pi}{8}}+\cos^4{\frac{\pi}{8}}\right\}\)
\(=2\left\{\left(\sin^2{\frac{\pi}{8}}\right)^2+\left(\cos^2{\frac{\pi}{8}}\right)^2\right\}\)
\(=2\left\{\left(\sin^2{\frac{\pi}{8}}+\cos^2{\frac{\pi}{8}}\right)^2-2\sin^2{\frac{\pi}{8}}\cos^2{\frac{\pi}{8}}\right\}\) ➜Note \(\because a^2+b^2=(a+b)^2-2ab\)

\(=2\left\{1^2-\frac{1}{2}\left(2\sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}}\right)^2\right\}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)

\(=2\left\{1-\frac{1}{2}\left(\sin{\frac{2\pi}{8}}\right)^2\right\}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=2\left\{1-\frac{1}{2}\left(\sin{\frac{\pi}{4}}\right)^2\right\}\)
\(=2\left\{1-\frac{1}{2}\left(\frac{1}{\sqrt{2}}\right)^2\right\}\) ➜Note \(\because \sin{\frac{\pi}{4}}=\frac{1}{\sqrt{2}}\)

\(=2\left\{1-\frac{1}{2}\times\frac{1}{2}\right\}\)
\(=2\left\{1-\frac{1}{4}\right\}\)
\(=2\times\frac{4-1}{4}\)
\(=\frac{3}{2}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sin^8{\theta}+\cos^8{\theta}\)
\(=(\sin^4{\theta})^2+(\cos^4{\theta})^2\)
\(=(\sin^4{\theta}+\cos^4{\theta})^2-2\sin^4{\theta}\cos^4{\theta}\) ➜Note \(\because a^2+b^2=(a+b)^2-2ab\)

\(=\{(\sin^2{\theta})^2+(\cos^2{\theta})^2\}^2-\frac{1}{8}(2\sin{\theta}\cos{\theta})^4\)
\(=\{(\sin^2{\theta}+\cos^2{\theta})^2-2\sin^2{\theta}\cos^2{\theta}\}^2-\frac{1}{8}(\sin{2\theta})^4\) ➜Note \(\because a^2+b^2=(a+b)^2-2ab\)
\(2\sin{A}\cos{A}=\sin{2A}\)

\(=\left\{1^2-\frac{1}{2}(2\sin{\theta}\cos{\theta})^2\right\}^2-\frac{1}{8}\sin^4{2\theta}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)

\(=\left\{1-\frac{1}{2}\sin^2{2\theta}\right\}^2-\frac{1}{8}\sin^4{2\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=1^2-2.1.\frac{1}{2}\sin^2{2\theta}+\frac{1}{4}\sin^4{2\theta}-\frac{1}{8}\sin^4{2\theta}\) ➜Note \(\because (a-b)^2=a^2-2ab+b^2\)

\(=1-\sin^2{2\theta}+\frac{2-1}{8}\sin^4{2\theta}\)
\(=1-\sin^2{2\theta}+\frac{1}{8}\sin^4{2\theta}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\sin{32\theta}\)
\(=\sin{2(16\theta)}\)
\(=2\sin{16\theta}\cos{16\theta}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=2\sin{2(8\theta)}\cos{16\theta}\)
\(=2\times2\sin{8\theta}\cos{8\theta}\cos{16\theta}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=4\times\sin{2(4\theta)}\cos{8\theta}\cos{16\theta}\)
\(=4\times2\sin{4\theta}\cos{4\theta}\cos{8\theta}\cos{16\theta}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=8\times\sin{2(2\theta)}\cos{4\theta}\cos{8\theta}\cos{16\theta}\)
\(=8\times2\sin{2\theta}\cos{2\theta}\cos{4\theta}\cos{8\theta}\cos{16\theta}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=16\times\sin{2\theta}\cos{2\theta}\cos{4\theta}\cos{8\theta}\cos{16\theta}\)
\(=16\times2\sin{\theta}\cos{\theta}\cos{2\theta}\cos{4\theta}\cos{8\theta}\cos{16\theta}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=32\sin{\theta}\cos{\theta}\cos{2\theta}\cos{4\theta}\cos{8\theta}\cos{16\theta}\)
\(=2^5\sin{\theta}\cos{\theta}\cos{2\theta}\cos{2^2\theta}\cos{2^3\theta}\cos{2^4\theta}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\cos{\theta}=\frac{1}{2}\left(x+\frac{1}{x}\right)\)
\((a)\)
\(L.S=\cos{2\theta}\)
\(=2\cos^2{\theta}-1\) ➜Note \(\because \cos{2A}=2\cos^2{A}-1\)

\(=2\left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\}^2-1\) ➜Note \(\because \cos{\theta}=\frac{1}{2}\left(x+\frac{1}{x}\right)\)

\(=2\left\{\frac{1}{4}\left(x+\frac{1}{x}\right)^2\right\}-1\)
\(=\frac{1}{2}\left(x+\frac{1}{x}\right)^2-1\)
\(=\frac{1}{2}\left(x^2+\frac{1}{x^2}+2x\times\frac{1}{x}\right)-1\) ➜Note \(\because (a+b)^2=a^2+b^2+2ab\)

\(=\frac{1}{2}\left(x^2+\frac{1}{x^2}+2\right)-1\)
\(=\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)+\frac{1}{2}\times2-1\)
\(=\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)+1-1\)
\(=\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
\(L.S=\cos{3\theta}\)
\(=4\cos^3{\theta}-3\cos{\theta}\) ➜Note \(\because \cos{3A}=4\cos^3{A}-3\cos{A}\)

\(=4\left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\}^3-3\times\frac{1}{2}\left(x+\frac{1}{x}\right)\) ➜Note \(\because \cos{\theta}=\frac{1}{2}\left(x+\frac{1}{x}\right)\)

\(=4\times\frac{1}{8}\left(x+\frac{1}{x}\right)^3-\frac{3}{2}\left(x+\frac{1}{x}\right)\)
\(=\frac{1}{2}\left(x+\frac{1}{x}\right)^3-\frac{3}{2}\left(x+\frac{1}{x}\right)\)
\(=\frac{1}{2}\left\{\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right\}\)
\(=\frac{1}{2}\left\{\left(x+\frac{1}{x}\right)^3-3x\times\frac{1}{x}\left(x+\frac{1}{x}\right)\right\}\)
\(=\frac{1}{2}\left\{x^3+\left(\frac{1}{x}\right)^3\right\}\)
\(=\frac{1}{2}\left(x^3+\frac{1}{x^3}\right)\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((c)\)
\(L.S=\cos{4\theta}\)
\(=\cos{2(2\theta)}\)
\(=2\cos^2{2\theta}-1\) ➜Note \(\because \cos{2A}=2\cos^2{A}-1\)

\(=2(\cos{2\theta})^2-1\)
\(=2(2\cos^2{\theta}-1)^2-1\) ➜Note \(\because \cos{2A}=2\cos^2{A}-1\)

\(=2\left\{2\times\frac{1}{4}\left(x+\frac{1}{x}\right)^2-1\right\}^2-1\) ➜Note \(\because \cos{\theta}=\frac{1}{2}\left(x+\frac{1}{x}\right)\)

\(=2\left\{\frac{1}{2}\left(x+\frac{1}{x}\right)^2-1\right\}^2-1\)
\(=2\times\frac{1}{4}\left\{\left(x+\frac{1}{x}\right)^2-2\right\}^2-1\)
\(=\frac{1}{2}\left\{\left(x+\frac{1}{x}\right)^2-2\right\}^2-1\)
\(=\frac{1}{2}\left\{\left(x+\frac{1}{x}\right)^2-2x\times\frac{1}{x}\right\}^2-1\)
\(=\frac{1}{2}\left\{(x)^2+\left(\frac{1}{x}\right)^2\right\}^2-1\) ➜Note \(\because (a+b)^2-2ab=a^2+b^2\)

\(=\frac{1}{2}\left\{x^2+\frac{1}{x^2}\right\}^2-1\)
\(=\frac{1}{2}\left\{\left(x^2+\frac{1}{x^2}\right)^2-2\right\}\)
\(=\frac{1}{2}\left\{\left(x^2+\frac{1}{x^2}\right)^2-2x^2\times\frac{1}{x^2}\right\}\)
\(=\frac{1}{2}\left\{(x^2)^2+\left(\frac{1}{x^2}\right)^2\right\}\) ➜Note \(\because (a+b)^2-2ab=a^2+b^2\)

\(=\frac{1}{2}\left(x^4+\frac{1}{x^4}\right)\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\tan{\theta}=\frac{1}{3}\) এবং \(\tan{\phi}=\frac{1}{7}\)
\(L.S=\sin{4\theta}\)
\(=\sin{2(2\theta)}\)
\(=2\sin{(2\theta)}\cos{(2\theta)}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=2\times\frac{2\tan{\theta}}{1+\tan^2{\theta}}\times\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}\) ➜Note \(\because \sin{2A}=\frac{2\tan{A}}{1+\tan^2{A}}\)
\(\cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(=\frac{4\times\frac{1}{3}}{1+\left(\frac{1}{3}\right)^2}\times\frac{1-\left(\frac{1}{3}\right)^2}{1+\left(\frac{1}{3}\right)^2}\) ➜Note \(\because \tan{\theta}=\frac{1}{3}\)

\(=\frac{\frac{4}{3}}{1+\frac{1}{9}}\times\frac{1-\frac{1}{9}}{1+\frac{1}{9}}\)
\(=\frac{4\times3}{9+1}\times\frac{9-1}{9+1}\) ➜Note উভয় ভগ্নাংশের লব ও হরকে \(9\) দ্বারা গুণ করে।

\(=\frac{12}{10}\times\frac{8}{10}\)
\(=\frac{6}{5}\times\frac{4}{5}\)
\(=\frac{24}{25}\)
আবার,
\(R.S=\cos{2\phi}\)
\(=\frac{1-\tan^2{\phi}}{1+\tan^2{\phi}}\) ➜Note \(\because \cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(=\frac{1-\left(\frac{1}{7}\right)^2}{1+\left(\frac{1}{7}\right)^2}\) ➜Note \(\because \tan{\phi}=\frac{1}{7}\)

\(=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}\)
\(=\frac{49-1}{49+1}\) ➜Note লব ও হরকে \(49\) দ্বারা গুণ করে।

\(=\frac{48}{50}\)
\(=\frac{24}{25}\)
\(\therefore L.S=R.S\)
(দেখানো হলো)

দেওয়া আছে,
\(\tan^2{\theta}=1+2\tan^2{\phi}\)
\(\Rightarrow 1+2\tan^2{\phi}=\tan^2{\theta}\)
\(\Rightarrow 2\tan^2{\phi}=\tan^2{\theta}-1\)
\(\therefore \tan^2{\phi}=\frac{\tan^2{\theta}-1}{2}\)
\(L.S=\cos{2\phi}\)
\(=\frac{1-\tan^2{\phi}}{1+\tan^2{\phi}}\) ➜Note \(\because \cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(=\frac{1-\frac{\tan^2{\theta}-1}{2}}{1+\frac{\tan^2{\theta}-1}{2}}\) ➜Note \(\because \tan^2{\phi}=\frac{\tan^2{\theta}-1}{2}\)

\(=\frac{2-(\tan^2{\theta}-1)}{2+\tan^2{\theta}-1}\) ➜Note লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=\frac{2-\tan^2{\theta}+1}{1+\tan^2{\theta}}\)
\(=\frac{3-\tan^2{\theta}}{1+\tan^2{\theta}}\)
\(=\frac{(1+\tan^2{\theta})+2(1-\tan^2{\theta})}{1+\tan^2{\theta}}\)
\(=\frac{1+\tan^2{\theta}}{1+\tan^2{\theta}}+2\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}\)
\(=1+2\cos{2\theta}\) ➜Note \(\because \frac{1-\tan^2{A}}{1+\tan^2{A}}=\cos{2A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(2\tan{\alpha}=3\tan{\beta}\)
\(\therefore \tan{\alpha}=\frac{3\tan{\beta}}{2}\)
\(L.S=\tan{(\alpha-\beta)}\)
\(=\frac{\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}}\) ➜Note \(\because \tan{(A-B)}=\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)

\(=\frac{\frac{3\tan{\beta}}{2}-\tan{\beta}}{1+\frac{3\tan{\beta}}{2}\times\tan{\beta}}\) ➜Note \(\because \tan{\alpha}=\frac{3\tan{\beta}}{2}\)

\(=\frac{\frac{3\tan{\beta}}{2}-\tan{\beta}}{1+\frac{3\tan^2{\beta}}{2}}\)
\(=\frac{3\tan{\beta}-2\tan{\beta}}{2+3\tan^2{\beta}}\) ➜Note লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=\frac{\tan{\beta}}{2+3\tan^2{\beta}}\)
\(=\frac{2\tan{\beta}}{4+6\tan^2{\beta}}\) ➜Note লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=\frac{2\tan{\beta}}{5(1+\tan^2{\beta})-(1-\tan^2{\beta})}\)
\(=\frac{\frac{2\tan{\beta}}{1+\tan^2{\beta}}}{\frac{5(1+\tan^2{\beta})}{1+\tan^2{\beta}}-\frac{1-\tan^2{\beta}}{1+\tan^2{\beta}}}\) ➜Note লব ও হরকে \((1+\tan^2{\beta})\) দ্বারা ভাগ করে।

\(=\frac{\sin{2\beta}}{5-\cos{2\beta}}\) ➜Note \(\because \frac{2\tan{A}}{1+\tan^2{A}}=\sin{2A}\)
এবং \(\frac{1-\tan^2{A}}{1+\tan^2{A}}=\cos{2A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\tan{\alpha}=2\tan{\beta}\)
\(\Rightarrow 2\tan{\beta}=\tan{\alpha}\)
\(\therefore \tan{\beta}=\frac{\tan{\alpha}}{2}\)
\(L.S=\tan{(\alpha+\beta)}\)
\(=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}\) ➜Note \(\because \tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)

\(=\frac{\tan{\alpha}+\frac{\tan{\alpha}}{2}}{1-\tan{\alpha}\times\frac{\tan{\alpha}}{2}}\) ➜Note \(\because \tan{\beta}=\frac{\tan{\alpha}}{2}\)

\(=\frac{\tan{\alpha}+\frac{\tan{\alpha}}{2}}{1-\frac{\tan^2{\alpha}}{2}}\)
\(=\frac{2\tan{\alpha}+\tan{\alpha}}{2-\tan^2{\alpha}}\) ➜Note লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=\frac{3\tan{\alpha}}{2-\tan^2{\alpha}}\)
\(=\frac{6\tan{\beta}}{4-2\tan^2{\beta}}\) ➜Note লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=\frac{6\tan{\beta}}{(1+\tan^2{\beta})+3(1-\tan^2{\beta})}\)
\(=\frac{3\frac{2\tan{\beta}}{1+\tan^2{\beta}}}{\frac{1+\tan^2{\beta}}{1+\tan^2{\beta}}+3\frac{1-\tan^2{\beta}}{1+\tan^2{\beta}}}\) ➜Note লব ও হরকে \((1+\tan^2{\beta})\) দ্বারা ভাগ করে।

\(=\frac{3\sin{2\alpha}}{1+3\cos{2\alpha}}\) ➜Note \(\because \frac{2\tan{A}}{1+\tan^2{A}}=\sin{2A}\)
এবং \(\frac{1-\tan^2{A}}{1+\tan^2{A}}=\cos{2A}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\tan{\theta}+2\tan{2\theta}+4\tan{4\theta}+8\cot{8\theta}\)
\(=\tan{\theta}+2\tan{2\theta}+4\tan{4\theta}+4\{2\cot{2(4\theta)}\}\)
\(=\tan{\theta}+2\tan{2\theta}+4\tan{4\theta}+4(\cot{4\theta}-\tan{4\theta})\) ➜Note \(\because 2\cot{2A}=\cot{A}-\tan{A}\)

\(=\tan{\theta}+2\tan{2\theta}+4\tan{4\theta}+4\cot{4\theta}-4\tan{4\theta}\)
\(=\tan{\theta}+2\tan{2\theta}+4\cot{4\theta}\)
\(=\tan{\theta}+2\tan{2\theta}+2\{2\cot{2(2\theta)}\}\)
\(=\tan{\theta}+2\tan{2\theta}+2(\cot{2\theta}-\tan{2\theta})\) ➜Note \(\because 2\cot{2A}=\cot{A}-\tan{A}\)

\(=\tan{\theta}+2\tan{2\theta}+2\cot{2\theta}-2\tan{2\theta}\)
\(=\tan{\theta}+2\cot{2\theta}\)
\(=\tan{\theta}+\cot{\theta}-\tan{\theta}\) ➜Note \(\because 2\cot{2A}=\cot{A}-\tan{A}\)

\(=\cot{\theta}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=(2\cos{\theta}-1)(2\cos{2\theta}-1)(2\cos{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\)
\(=\frac{1}{2\cos{\theta}+1}\times\{(2\cos{\theta}+1)(2\cos{\theta}-1)\}(2\cos{2\theta}-1)(2\cos{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\)
\(=\frac{1}{2\cos{\theta}+1}\times(4\cos^2{\theta}-1)(2\cos{2\theta}-1)(2\cos{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\) ➜Note \(\because (a+b)(a-b)=a^2-b^2\)

\(=\frac{1}{2\cos{\theta}+1}\times(2\times2\cos^2{\theta}-1)(2\cos{2\theta}-1)(2\cos{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\)
\(=\frac{1}{2\cos{\theta}+1}\times\{2(1+\cos{2\theta})-1\}(2\cos{2\theta}-1)(2\cos{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{1}{2\cos{\theta}+1}\times\{2+2\cos{2\theta}-1\}(2\cos{2\theta}-1)(2\cos{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\)
\(=\frac{1}{2\cos{\theta}+1}\times\{(2\cos{2\theta}+1)(2\cos{2\theta}-1)\}(2\cos{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\)
\(=\frac{1}{2\cos{\theta}+1}\times(4\cos^2{2\theta}-1)(2\cos{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\) ➜Note \(\because (a+b)(a-b)=a^2-b^2\)

\(=\frac{1}{2\cos{\theta}+1}\times\{2(2\cos^2{2\theta})-1\}(2\cos{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\)
\(=\frac{1}{2\cos{\theta}+1}\times\{2(1+\cos{4\theta})-1\}(2\cos{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{1}{2\cos{\theta}+1}\times\{2+2\cos{2^2\theta}-1\}(2\cos{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\)
\(=\frac{1}{2\cos{\theta}+1}\times\{(2\cos{2^2\theta}+1)(2\cos{2^2\theta}-1)\}.....(2\cos{2^{n-1}\theta}-1)\)
\(=\frac{1}{2\cos{\theta}+1}\times(4\cos^2{2^2\theta}-1).....(2\cos{2^{n-1}\theta}-1)\) ➜Note \(\because (a+b)(a-b)=a^2-b^2\)

\(=\frac{1}{2\cos{\theta}+1}\times\{2(2\cos^2{2^2\theta})-1\}.....(2\cos{2^{n-1}\theta}-1)\)
\(=\frac{1}{2\cos{\theta}+1}\times\{2(1+\cos{2^3\theta})-1\}.....(2\cos{2^{n-1}\theta}-1)\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{1}{2\cos{\theta}+1}\times(2+2\cos{2^3\theta}-1).....(2\cos{2^{n-1}\theta}-1)\)
\(=\frac{1}{2\cos{\theta}+1}\times(2\cos{2^3\theta}+1).....(2\cos{2^{n-1}\theta}-1)\)
\(..........................................\)
\(..........................................\)
\(=\frac{1}{2\cos{\theta}+1}\times\{(2\cos{2^{n-1}\theta}+1)(2\cos{2^{n-1}\theta}-1)\}\)
\(=\frac{1}{2\cos{\theta}+1}\times(4\cos^2{2^{n-1}\theta}-1)\) ➜Note \(\because (a+b)(a-b)=a^2-b^2\)

\(=\frac{1}{2\cos{\theta}+1}\times\{2(2\cos^2{2^{n-1}\theta})-1\}\)
\(=\frac{1}{2\cos{\theta}+1}\times\{2(1+\cos{2^{n-1+1}\theta})-1\}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)

\(=\frac{1}{2\cos{\theta}+1}\times(2+2\cos{2^{n}\theta}-1\}\)
\(=\frac{1}{2\cos{\theta}+1}\times(2\cos{2^{n}\theta}+1)\)
\(=\frac{2\cos{2^{n}\theta}+1}{2\cos{\theta}+1}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

\(L.S=\frac{\cos{(45^{o}+A)}}{\cos{(45^{o}-A)}}\)
\(=\frac{2\cos^2{(45^{o}+A)}}{2\cos{(45^{o}+A)}\cos{(45^{o}-A)}}\) ➜Note লব ও হরকে \(2\cos{(45^{o}+A)}\) দ্বারা গুণ করে।

\(=\frac{1+\cos{2(45^{o}+A)}}{\cos{(45^{o}+A-45^{o}+A)}+\cos{(45^{o}+A+45^{o}-A)}}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)
এবং \(2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)

\(=\frac{1+\cos{(90^{o}+2A)}}{\cos{2A}+\cos{90^{o}}}\)
\(=\frac{1+\cos{(90^{o}\times1+2A)}}{\cos{2A}+0}\) ➜Note \(\because 90^{o}+2A=90^{o}\times1+2A\)
এবং \(\cos{90^{o}}=0\)

\(=\frac{1-\sin{2A}}{\cos{2A}}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ঋনাত্মক।
আবার, \(90^{o}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
তাই কোসাইন অনুপাতের পরিবর্তন হয়ে সাইন হয়েছে।

\(=\frac{1}{\cos{2A}}-\frac{\sin{2A}}{\cos{2A}}\)
\(=\sec{2A}-\tan{2A}\) ➜Note \(\because \frac{1}{\cos{P}}=\sec{P}\)
এবং \(\frac{\sin{P}}{\cos{P}}=\tan{P}\)

\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\cos{2\alpha}=\frac{3\cos{2\beta}-1}{3-\cos{2\beta}}\)
\(\Rightarrow \frac{1}{\cos{2\alpha}}=\frac{3-\cos{2\beta}}{3\cos{2\beta}-1}\) ➜Note ব্যাস্তকরণ করে।

\(\Rightarrow \frac{1-\cos{2\alpha}}{1+\cos{2\alpha}}=\frac{3-\cos{2\beta}-3\cos{2\beta}+1}{3-\cos{2\beta}+3\cos{2\beta}-1}\) ➜Note বিয়োজন-যোজন করে।

\(\Rightarrow \tan^2{\alpha}=\frac{4-4\cos{2\beta}}{2+2\cos{2\beta}}\) ➜Note \(\because \frac{1-\cos{2A}}{1+\cos{2A}}=\tan^2{A}\)

\(\Rightarrow \tan^2{\alpha}=\frac{4(1-\cos{2\beta})}{2(1+\cos{2\beta})}\)
\(\Rightarrow \tan^2{\alpha}=\frac{2(1-\cos{2\beta})}{1+\cos{2\beta}}\)
\(\Rightarrow \tan^2{\alpha}=2\frac{1-\cos{2\beta}}{1+\cos{2\beta}}\)
\(\Rightarrow \tan^2{\alpha}=2\tan^2{\beta}\) ➜Note \(\because \frac{1-\cos{2A}}{1+\cos{2A}}=\tan^2{A}\)

\(\Rightarrow \tan{\alpha}=\sqrt{2\tan^2{\beta}}\)
\(\therefore \tan{\alpha}=\sqrt{2}\tan{\beta}\) ➜Note \(\because \) \(\alpha\) ও \(\beta\) কোণদ্বয় ধনাত্মক ও সূক্ষ্ণ।

(দেখানো হলো)

দেওয়া আছে,
\(\tan{\theta}=\frac{\tan{x}+\tan{y}}{1+\tan{x}\tan{y}}\)
\(\Rightarrow \tan{\theta}=\frac{\frac{\sin{x}}{\cos{x}}+\frac{\sin{y}}{\cos{y}}}{1+\frac{\sin{x}}{\cos{x}}\times\frac{\sin{y}}{\cos{y}}}\) ➜Note \(\because \tan{P}=\frac{\sin{P}}{\cos{P}}\)

\(\Rightarrow \tan{\theta}=\frac{\frac{\sin{x}}{\cos{x}}+\frac{\sin{y}}{\cos{y}}}{1+\frac{\sin{x}\sin{y}}{\cos{x}\cos{y}}}\)
\(\Rightarrow \tan{\theta}=\frac{\sin{x}\cos{y}+\cos{x}\sin{y}}{\cos{x}\cos{y}+\sin{x}\sin{y}}\) ➜Note লব ও হরকে \(\cos{x}\cos{y}\) দ্বারা গুণ করে।

\(\therefore \tan{\theta}=\frac{\sin{(x+y)}}{\cos{(x-y)}}\) ➜Note \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
এবং \(\cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)

\(L.S=\sin{2\theta}\)
\(=\frac{2\tan{\theta}}{1+\tan^2{\theta}}\) ➜Note \(\because \sin{2A}=\frac{2\tan{A}}{1+\tan^2{A}}\)

\(=\frac{2\frac{\sin{(x+y)}}{\cos{(x-y)}}}{1+\left\{\frac{\sin{(x+y)}}{\cos{(x-y)}}\right\}^2}\) ➜Note \(\because \tan{\theta}=\frac{\sin{(x+y)}}{\cos{(x-y)}}\)

\(=\frac{2\frac{\sin{(x+y)}}{\cos{(x-y)}}}{1+\frac{\sin^2{(x+y)}}{\cos^2{(x-y)}}}\)
\(=\frac{2\sin{(x+y)}\cos{(x-y)}}{\cos^2{(x-y)}+\sin^2{(x+y)}}\) ➜Note লব ও হরকে \(\cos^2{(x-y)}\) দ্বারা গুণ করে।

\(=\frac{\sin{(x+y+x-y)}+\sin{(x+y-x+y)}}{1-\sin^2{(x-y)}+\sin^2{(x+y)}}\) ➜Note \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
এবং \(\cos^2{A}=1-\sin^2{A}\)

\(=\frac{\sin{2x}+\sin{2y}}{1+\{\sin^2{(x+y)}-\sin^2{(x-y)}\}}\)
\(=\frac{\sin{2x}+\sin{2y}}{1+\sin{(x+y+x-y)}\sin{(x+y-x+y)}}\) ➜Note \(\because \sin^2{A}-\sin^2{B}=\sin{(A+B)}\sin{(A-B)}\)

\(=\frac{\sin{2x}+\sin{2y}}{1+\sin{2x}\sin{2y}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\tan{\beta}=\frac{1}{3}\)
প্রদত্ত রাশি,
\(=\sin{2\beta}\) ➜Note \(\because \tan{P}=\frac{\sin{P}}{\cos{P}}\)

\(=\frac{2\tan{\beta}}{1+\tan^2{\beta}}\) ➜Note \(\because \sin{2P}=\frac{2\tan{P}}{1+\tan^2{P}}\)

\(=\frac{2\times\frac{1}{3}}{1+\left(\frac{1}{3}\right)^2}\) ➜Note \(\because \tan{\beta}=\frac{1}{3}\)

\(=\frac{\frac{2}{3}}{1+\frac{1}{9}}\)
\(=\frac{2\times3}{9+1}\) ➜Note লব ও হরকে \(9\) দ্বারা গুণ করে।

\(=\frac{6}{10}\)
\(=\frac{3}{5}\)
ইহাই নির্ণেয় মান।

দেওয়া আছে,
\(\cot{x}=\frac{1}{9}\)
\(\Rightarrow \frac{1}{\cot{x}}=9\) ➜Note ব্যাস্তকরণ করে।

\(\therefore \tan{x}=9\) ➜Note \(\because \frac{1}{\cot{A}}=\tan{A}\)

প্রদত্ত রাশি,
\(=\sec{2x}\)
\(=\frac{1}{\cos{2x}}\) ➜Note \(\because \sec{P}=\frac{1}{\cos{P}}\)

\(=\frac{1}{\frac{1-\tan^2{x}}{1+\tan^2{x}}}\) ➜Note \(\because \cos{2P}=\frac{1-\tan^2{P}}{1+\tan^2{P}}\)

\(=\frac{1+\tan^2{x}}{1-\tan^2{x}}\)
\(=\frac{1+9^2}{1-9^2}\) ➜Note \(\because \tan{x}=9\)

\(=\frac{82}{-80}\)
\(=-\frac{41}{40}\)
ইহাই নির্ণেয় মান।

দেওয়া আছে,
\(\cot{\theta}=2\)
\(\Rightarrow \frac{1}{\cot{\theta}}=\frac{1}{2}\) ➜Note ব্যাস্তকরণ করে।

\(\therefore \tan{\theta}=\frac{1}{2}\) ➜Note \(\because \frac{1}{\cot{A}}=\tan{A}\)

প্রদত্ত রাশি,
\(=10\sin{2\theta}-6\tan{2\theta}\)
\(=10\frac{2\tan{\theta}}{1+\tan^2{\theta}}-6\frac{2\tan{\theta}}{1-\tan^2{\theta}}\) ➜Note \(\because \sin{2P}=\frac{2\tan{P}}{1+\tan^2{P}}\)
\(\tan{2P}=\frac{2\tan{P}}{1-\tan^2{P}}\)

\(=10\frac{2\times\frac{1}{2}}{1+\left(\frac{1}{2}\right)^2}-6\frac{2\times\frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}\) ➜Note \(\because \tan{\theta}=\frac{1}{2}\)

\(=10\frac{1}{1+\frac{1}{4}}-6\frac{1}{1-\frac{1}{4}}\)
\(=10\frac{1}{4+1}-6\frac{1}{4-1}\) ➜Note লব ও হরকে \(4\) দ্বারা গুণ করে।

\(=10\times\frac{1}{5}-6\times\frac{1}{3}\)
\(=2-2\)
\(=0\)
ইহাই নির্ণেয় মান।

দেওয়া আছে,
\(\theta=\cos^{-1}\frac{1}{3}\)
\(\therefore \cos{\theta}=\frac{1}{3}\)
প্রদত্ত রাশি,
\(=\cos{3\theta}\)
\(=4\cos^3{\theta}-3\cos{\theta}\) ➜Note \(\because \cos{3P}=4\cos^3{P}-3\cos{P}\)

\(=4\left(\frac{1}{3}\right)^3-3\times\frac{1}{3}\) ➜Note \(\because \cos{\theta}=\frac{1}{3}\)

\(=4\times\frac{1}{27}-1\)
\(=\frac{4}{27}-1\)
\(=\frac{4-27}{27}\)
\(=\frac{-23}{27}\)
\(=-\frac{23}{27}\)
ইহাই নির্ণেয় মান।

দেওয়া আছে,
\(\tan{\alpha}=\frac{1}{7}\) এবং \(\tan{\beta}=\frac{1}{3}\)
\(L.S=\cos{2\alpha}\)
\(=\frac{1-\tan^2{\alpha}}{1+\tan^2{\alpha}}\) ➜Note \(\because \cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(=\frac{1-\left(\frac{1}{7}\right)^2}{1+\left(\frac{1}{7}\right)^2}\) ➜Note \(\because \tan{\alpha}=\frac{1}{7}\)

\(=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}\)
\(=\frac{49-1}{49+1}\) ➜Note লব ও হরকে \(49\) দ্বারা গুণ করে।

\(=\frac{48}{50}\)
\(=\frac{24}{25}\)
আবার,
\(R.S=\sin{4\beta}\)
\(=\sin{2(2\beta)}\)
\(=2\sin{(2\beta)}\cos{(2\beta)}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=2\times\frac{2\tan{\beta}}{1+\tan^2{\beta}}\times\frac{1-\tan^2{\beta}}{1+\tan^2{\beta}}\) ➜Note \(\because \sin{2A}=\frac{2\tan{A}}{1+\tan^2{A}}\)
\(\cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(=\frac{4\times\frac{1}{3}}{1+\left(\frac{1}{3}\right)^2}\times\frac{1-\left(\frac{1}{3}\right)^2}{1+\left(\frac{1}{3}\right)^2}\) ➜Note \(\because \tan{\beta}=\frac{1}{3}\)

\(=\frac{\frac{4}{3}}{1+\frac{1}{9}}\times\frac{1-\frac{1}{9}}{1+\frac{1}{9}}\)
\(=\frac{4\times3}{9+1}\times\frac{9-1}{9+1}\) ➜Note উভয় ভগ্নাংশের লব ও হরকে \(9\) দ্বারা গুণ করে।

\(=\frac{12}{10}\times\frac{8}{10}\)
\(=\frac{6}{5}\times\frac{4}{5}\)
\(=\frac{24}{25}\)
\(\therefore L.S=R.S\)
(দেখানো হলো)

দেওয়া আছে,
\(\tan{\theta}=\frac{a}{b}\)
\(L.S=\sin{4\theta}\)
\(=\sin{2(2\theta)}\)
\(=2\sin{(2\theta)}\cos{(2\theta)}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(=2\times\frac{2\tan{\theta}}{1+\tan^2{\theta}}\times\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}\) ➜Note \(\because \sin{2A}=\frac{2\tan{A}}{1+\tan^2{A}}\)
\(\cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(=\frac{4\times\frac{a}{b}}{1+\left(\frac{a}{b}\right)^2}\times\frac{1-\left(\frac{a}{b}\right)^2}{1+\left(\frac{a}{b}\right)^2}\) ➜Note \(\because \tan{\theta}=\frac{a}{b}\)

\(=\frac{\frac{4a}{b}}{1+\frac{a^2}{b^2}}\times\frac{1-\frac{a^2}{b^2}}{1+\frac{a^2}{b^2}}\)
\(=\frac{4ab}{b^2+a^2}\times\frac{b^2-a^2}{b^2+a^2}\) ➜Note উভয় ভগ্নাংশের লব ও হরকে \(b^2\) দ্বারা গুণ করে।

\(=\frac{4ab(b^2-a^2)}{(b^2+a^2)^2}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\angle{E}+\angle{F}=65^{o}, \ \angle{F}-\angle{E}=25^{o}\)
ধরি,
\(\angle{E}+\angle{F}=65^{o}........(1)\)
\(\angle{F}-\angle{E}=25^{o} ........(2)\)
\((1)\) হতে \((2)\) বিয়োগ করে।
\(\angle{E}+\angle{F}-\angle{F}+\angle{E}=65^{o}-25^{o}\)
\(\Rightarrow 2\angle{E}=40^{o}\)
\(\Rightarrow \angle{E}=\frac{40^{o}}{2}\)
\(\therefore \angle{E}=20^{o}\)
\(L.S=\tan{\angle{E}}\tan{2\angle{E}}\tan{3\angle{E}}\tan{4\angle{E}}\)
\(=\tan{20^{o}}\tan{2\times20^{o}}\tan{3\times20^{o}}\tan{4\times20^{o}}\)
\(=\tan{20^{o}}\tan{40^{o}}\tan{60^{o}}\tan{80^{o}}\)
\(=\tan{20^{o}}\tan{40^{o}}\times\sqrt{3}\tan{80^{o}}\) ➜Note \(\because \tan{60^{o}}=\sqrt{3}\)

\(=\sqrt{3}\tan{20^{o}}\tan{40^{o}}\tan{80^{o}}\)
\(=\sqrt{3}\tan{20^{o}}\tan{(60^{o}-20^{o})}\tan{(60^{o}+20^{o})}\) ➜Note \(\because 40^{o}=60^{o}-20^{o}\)
এবং \(80^{o}=60^{o}+20^{o}\)

\(=\sqrt{3}\tan{20^{o}}\times\frac{\tan{60^{o}}-\tan{20^{o}}}{1+\tan{60^{o}}\tan{20^{o}}}\times\frac{\tan{60^{o}}+\tan{20^{o}}}{1-\tan{60^{o}}\tan{20^{o}}}\) ➜Note \(\because \tan{(A-B)}=\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)
এবং \(\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)

\(=\sqrt{3}\tan{20^{o}}\times\frac{\sqrt{3}-\tan{20^{o}}}{1+\sqrt{3}\tan{20^{o}}}\times\frac{\sqrt{3}+\tan{20^{o}}}{1-\sqrt{3}\tan{20^{o}}}\) ➜Note \(\because \tan{60^{o}}=\sqrt{3}\)
\(=\sqrt{3}\tan{20^{o}}\times\frac{(\sqrt{3})^2-\tan^2{20^{o}}}{1^2-(\sqrt{3}\tan{20^{o}})^2}\) ➜Note \(\because (a-b)(a+b)=a^2-b^2\)
\(=\sqrt{3}\tan{20^{o}}\times\frac{3-\tan^2{20^{o}}}{1-3\tan^2{20^{o}}}\)
\(=\sqrt{3}\frac{3\tan{20^{o}}-\tan^3{20^{o}}}{1-3\tan^2{20^{o}}}\)
\(=\sqrt{3}\tan{(3\times20^{o})}\) ➜Note \(\because \frac{3\tan{A}-\tan^3{A}}{1-3\tan^2{A}}=\tan{3A}\)
\(=\sqrt{3}\tan{60^{o}}\)
\(=\sqrt{3}\times\sqrt{3}\) ➜Note \(\because \tan{60^{o}}=\sqrt{3}\)
\(=3\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(A=20^{o}, \ B=2A, \ C=3A, \ D=4A\)
\(\Rightarrow A=20^{o}, \ B=2\times20^{o}, \ C=3\times20^{o}, \ D=4\times20^{o}\)
\(\therefore A=20^{o}, \ B=40^{o}, \ C=60^{o}, \ D=80^{o}\)
\(L.S=\tan{A}\tan{B}\tan{C}\tan{D}\)
\(=\tan{20^{o}}\tan{40^{o}}\tan{60^{o}}\tan{80^{o}}\) ➜Note \(\because A=20^{o}, \ B=40^{o}, \ C=60^{o}, \ D=80^{o}\)

\(=\tan{20^{o}}\tan{40^{o}}\times\sqrt{3}\tan{80^{o}}\) ➜Note \(\because \tan{60^{o}}=\sqrt{3}\)

\(=\sqrt{3}\tan{20^{o}}\tan{40^{o}}\tan{80^{o}}\)
\(=\sqrt{3}\tan{20^{o}}\tan{(60^{o}-20^{o})}\tan{(60^{o}+20^{o})}\) ➜Note \(\because 40^{o}=60^{o}-20^{o}\)
এবং \(80^{o}=60^{o}+20^{o}\)

\(=\sqrt{3}\tan{20^{o}}\times\frac{\tan{60^{o}}-\tan{20^{o}}}{1+\tan{60^{o}}\tan{20^{o}}}\times\frac{\tan{60^{o}}+\tan{20^{o}}}{1-\tan{60^{o}}\tan{20^{o}}}\) ➜Note \(\because \tan{(A-B)}=\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)
এবং \(\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)

\(=\sqrt{3}\tan{20^{o}}\times\frac{\sqrt{3}-\tan{20^{o}}}{1+\sqrt{3}\tan{20^{o}}}\times\frac{\sqrt{3}+\tan{20^{o}}}{1-\sqrt{3}\tan{20^{o}}}\) ➜Note \(\because \tan{60^{o}}=\sqrt{3}\)
\(=\sqrt{3}\tan{20^{o}}\times\frac{(\sqrt{3})^2-\tan^2{20^{o}}}{1^2-(\sqrt{3}\tan{20^{o}})^2}\) ➜Note \(\because (a-b)(a+b)=a^2-b^2\)
\(=\sqrt{3}\tan{20^{o}}\times\frac{3-\tan^2{20^{o}}}{1-3\tan^2{20^{o}}}\)
\(=\sqrt{3}\frac{3\tan{20^{o}}-\tan^3{20^{o}}}{1-3\tan^2{20^{o}}}\)
\(=\sqrt{3}\tan{(3\times20^{o})}\) ➜Note \(\because \frac{3\tan{A}-\tan^3{A}}{1-3\tan^2{A}}=\tan{3A}\)
\(=\sqrt{3}\tan{60^{o}}\)
\(=\sqrt{3}\times\sqrt{3}\) ➜Note \(\because \tan{60^{o}}=\sqrt{3}\)
\(=3\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\cos{\alpha}+\cos{\beta}+\cos{\gamma}=0\)
\(\therefore \cos{\alpha}+\cos{\beta}=-\cos{\gamma}\)
\(L.S=\cos{3\alpha}+\cos{3\beta}+\cos{3\gamma}\)
\(=4\cos^3{\alpha}-3\cos{\alpha}+4\cos^3{\beta}-3\cos{\beta}+4\cos^3{\gamma}-3\cos{\gamma}\) ➜Note \(\because \cos{3A}=4\cos^3{A}-3\cos{A}\)

\(=4(\cos^3{\alpha}+\cos^3{\beta})+4\cos^3{\gamma}-3(\cos{\alpha}+\cos{\beta}+\cos{\gamma})\)
\(=4\{(\cos{\alpha}+\cos{\beta})^3-3\cos{\alpha}\cos{\beta}(\cos{\alpha}+\cos{\beta})\}+4\cos^3{\gamma}-3(0)\) ➜Note \(\because a^3+b^3=(a+b)^3-3ab(a+b)\)
এবং \(\cos{\alpha}+\cos{\beta}+\cos{\gamma}=0\)

\(=4\{(-\cos{\gamma})^3-3\cos{\alpha}\cos{\beta}(-\cos{\gamma})\}+4\cos^3{\gamma}\) ➜Note \(\because \cos{\alpha}+\cos{\beta}=-\cos{\gamma}\)

\(=4\{-\cos^3{\gamma}+3\cos{\alpha}\cos{\beta}\cos{\gamma}\}+4\cos^3{\gamma}\)
\(=-4\cos^3{\gamma}+12\cos{\alpha}\cos{\beta}\cos{\gamma}+4\cos^3{\gamma}\)
\(=12\cos{\alpha}\cos{\beta}\cos{\gamma}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)

দেওয়া আছে,
\(cosec \ {2A}+cosec \ {2B}+cosec \ {2C}=0\)
\(L.S=\tan{A}+\tan{B}+\tan{C}+\cot{A}+\cot{B}+\cot{C}\)
\(=\left(\tan{A}+\cot{A}\right)+\left(\tan{B}+\cot{B}\right)+\left(\tan{C}+\cot{C}\right)\)
\(=\left(\frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}\right)+\left(\frac{\sin{B}}{\cos{B}}+\frac{\cos{B}}{\sin{B}}\right)+\left(\frac{\sin{C}}{\cos{C}}+\frac{\cos{C}}{\sin{C}}\right)\) ➜Note \(\because \tan{P}=\frac{\sin{P}}{\cos{P}}\)
এবং \(\cot{P}=\frac{\cos{P}}{\sin{P}}\)

\(=\frac{\sin^2{A}+\cos^2{A}}{\sin{A}\cos{A}}+\frac{\sin^2{B}+\cos^2{B}}{\sin{B}\cos{B}}+\frac{\sin^2{C}+\cos^2{C}}{\sin{C}\cos{C}}\)
\(=\frac{1}{\sin{A}\cos{A}}+\frac{1}{\sin{B}\cos{B}}+\frac{1}{\sin{C}\cos{C}}\) ➜Note \(\because \sin^2{P}+\cos^2{P}=1\)

\(=\frac{2}{2\sin{A}\cos{A}}+\frac{2}{2\sin{B}\cos{B}}+\frac{2}{2\sin{C}\cos{C}}\) ➜Note প্রতিটি ভগ্নাংশের লব ও হরকে \(2\) দ্বারা গুণ করে।

\(=2\frac{1}{\sin{2A}}+2\frac{1}{\sin{2B}}+2\frac{1}{\sin{2C}}\) ➜Note \(\because 2\sin{P}\cos{P}=\sin{2P}\)

\(=2cosec \ {2A}+2cosec \ {2B}+2cosec \ {2C}\) ➜Note \(\because \frac{1}{\sin{P}}=cosec \ {P}\)

\(=2(cosec \ {2A}+cosec \ {2B}+cosec \ {2C})\)
\(=2\times0\) ➜Note \(\because cosec \ {2A}+cosec \ {2B}+cosec \ {2C}=0\)

\(=0\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\tan{\theta}\tan{\phi}=\sqrt{\frac{a-b}{a+b}}\)
\(\Rightarrow \tan^2{\theta}\tan^2{\phi}=\frac{a-b}{a+b}\) ➜Note উভয় পার্শে বর্গ করে।

\(L.S=(a-b\cos{2\theta})(a-b\cos{2\phi})\)
\(=\left(a-b\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}\right)\times\left(a-b\frac{1-\tan^2{\phi}}{1+\tan^2{\phi}}\right)\) ➜Note \(\because \cos{2P}=\frac{1-\tan^2{P}}{1+\tan^2{P}}\)

\(=\left(a-\frac{b-b\tan^2{\theta}}{1+\tan^2{\theta}}\right)\times\left(a-\frac{b-b\tan^2{\phi}}{1+\tan^2{\phi}}\right)\)
\(=\left(\frac{a+a\tan^2{\theta}-b+b\tan^2{\theta}}{1+\tan^2{\theta}}\right)\times\left(\frac{a+a\tan^2{\phi}-b+b\tan^2{\phi}}{1+\tan^2{\phi}}\right)\)
\(=\left\{\frac{(a-b)+(a+b)\tan^2{\theta}}{1+\tan^2{\theta}}\right\}\times\left\{\frac{(a-b)+(a+b)\tan^2{\phi}}{1+\tan^2{\phi}}\right\}\)
\(=\left\{(a+b)\frac{\frac{a-b}{a+b}+\tan^2{\theta}}{1+\tan^2{\theta}}\right\}\times\left\{(a+b)\frac{\frac{a-b}{a+b}+\tan^2{\phi}}{1+\tan^2{\phi}}\right\}\)
\(=\left\{(a+b)\frac{\tan^2{\theta}\tan^2{\phi}+\tan^2{\theta}}{1+\tan^2{\theta}}\right\}\times\left\{(a+b)\frac{\tan^2{\theta}\tan^2{\phi}+\tan^2{\phi}}{1+\tan^2{\phi}}\right\}\) ➜Note \(\because \tan^2{\theta}\tan^2{\phi}=\frac{a-b}{a+b}\)

\(=\left\{(a+b)\frac{\tan^2{\theta}(1+\tan^2{\phi})}{1+\tan^2{\theta}}\right\}\times\left\{(a+b)\frac{\tan^2{\phi}(1+\tan^2{\theta})}{1+\tan^2{\phi}}\right\}\)
\(=\left\{(a+b)\tan^2{\theta}\right\}\times\left\{(a+b)\tan^2{\phi}\right\}\)
\(=(a+b)^2\tan^2{\theta}\tan^2{\phi}\)
\(=(a+b)^2\times\frac{a-b}{a+b}\) ➜Note \(\because \tan^2{\theta}\tan^2{\phi}=\frac{a-b}{a+b}\)

\(=(a+b)(a-b)\)
\(=a^2-b^2\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)

দেওয়া আছে,
\(\tan{\beta}=\frac{\sin{2\alpha}}{5+\cos{2\alpha}}\)
\(\Rightarrow \tan{\beta}=\frac{2\sin{\alpha}\cos{\alpha}}{4+(1+\cos{2\alpha})}\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)

\(\Rightarrow \tan{\beta}=\frac{2\sin{\alpha}\cos{\alpha}}{4+2\cos^2{\alpha}}\) ➜Note \(\because 1+\cos{2\alpha}=2\cos^2{A}\)

\(\Rightarrow \tan{\beta}=\frac{2\sin{\alpha}\cos{\alpha}}{2(2+\cos^2{\alpha})}\)
\(\therefore \tan{\beta}=\frac{\sin{\alpha}\cos{\alpha}}{2+\cos^2{\alpha}}\)
এখন, \(\tan{(\alpha-\beta)}=\frac{\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}}\)
\(=\frac{\frac{\sin{\alpha}}{\cos{\alpha}}-\frac{\sin{\alpha}\cos{\alpha}}{2+\cos^2{\alpha}}}{1+\frac{\sin{\alpha}}{\cos{\alpha}}\times\frac{\sin{\alpha}\cos{\alpha}}{2+\cos^2{\alpha}}}\) ➜Note \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
এবং \(\tan{\beta}=\frac{\sin{\alpha}\cos{\alpha}}{2+\cos^2{\alpha}}\)

\(=\frac{\frac{\sin{\alpha}}{\cos{\alpha}}-\frac{\sin{\alpha}\cos{\alpha}}{2+\cos^2{\alpha}}}{1+\frac{\sin^2{\alpha}\cos{\alpha}}{\cos{\alpha}(2+\cos^2{\alpha})}}\)
\(=\frac{2\sin{\alpha}+\sin{\alpha}\cos^2{\alpha}-\sin{\alpha}\cos^2{\alpha}}{2\cos{\alpha}+\cos^3{\alpha}+\sin^2{\alpha}\cos{\alpha}}\) ➜Note লব ও হরকে \(\cos{\alpha}(2+\cos^2{\alpha})\) দ্বারা গুণ করে।

\(=\frac{2\sin{\alpha}}{2\cos{\alpha}+\cos{\alpha}(\sin^2{\alpha}+\cos^2{\alpha})}\)
\(=\frac{2\sin{\alpha}}{2\cos{\alpha}+\cos{\alpha}.1}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)

\(=\frac{2\sin{\alpha}}{2\cos{\alpha}+\cos{\alpha}}\)
\(=\frac{2\sin{\alpha}}{3\cos{\alpha}}\)
\(=\frac{2}{3}\times\frac{\sin{\alpha}}{\cos{\alpha}}\)
\(=\frac{2}{3}\tan{\alpha}\) ➜Note \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)

\(\therefore \tan{(\alpha-\beta)}=\frac{2}{3}\tan{\alpha}\)
\(\therefore 3\tan{(\alpha-\beta)}=2\tan{\alpha}\)
(দেখানো হলো)

দেওয়া আছে,
\(\cos{A}+\cos{B}+\cos{C}=0\)
\(\therefore \cos{A}+\cos{B}=-\cos{C}\)
\(L.S=\cos{3A}+\cos{3B}+\cos{3C}\)
\(=4\cos^3{A}-3\cos{A}+4\cos^3{B}-3\cos{B}+4\cos^3{C}-3\cos{C}\) ➜Note \(\because \cos{3A}=4\cos^3{A}-3\cos{A}\)

\(=4(\cos^3{A}+\cos^3{B})+4\cos^3{C}-3(\cos{A}+\cos{B}+\cos{C})\)
\(=4\{(\cos{A}+\cos{B})^3-3\cos{A}\cos{B}(\cos{A}+\cos{B})\}+4\cos^3{C}-3(0)\) ➜Note \(\because a^3+b^3=(a+b)^3-3ab(a+b)\)
এবং \(\cos{A}+\cos{B}+\cos{C}=0\)

\(=4\{(-\cos{C})^3-3\cos{A}\cos{B}(-\cos{C})\}+4\cos^3{C}\) ➜Note \(\because \cos{A}+\cos{B}=-\cos{C}\)

\(=4\{-\cos^3{C}+3\cos{A}\cos{B}\cos{C}\}+4\cos^3{C}\)
\(=-4\cos^3{C}+12\cos{A}\cos{B}\cos{C}+4\cos^3{C}\)
\(=12\cos{A}\cos{B}\cos{C}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)

দেওয়া আছে,
\(a\cos{\alpha}+b\sin{\alpha}=a\cos{\beta}+b\sin{\beta}\)
\(\Rightarrow a\cos{\alpha}-a\cos{\beta}=b\sin{\beta}-b\sin{\alpha}\)
\(\Rightarrow a(\cos{\alpha}-\cos{\beta})=b(\sin{\beta}-\sin{\alpha})\)
\(\Rightarrow a\times2\sin{\left(\frac{\alpha+\beta}{2}\right)}\sin{\left(\frac{\beta-\alpha}{2}\right)}=b\times2\cos{\left(\frac{\alpha+\beta}{2}\right)}\sin{\left(\frac{\beta-\alpha}{2}\right)}\) ➜Note \(\because \cos{D}-\cos{C}=2\sin{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
এবং \(\sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)

\(\Rightarrow a\sin{\left(\frac{\alpha+\beta}{2}\right)}=b\cos{\left(\frac{\alpha+\beta}{2}\right)}\)
\(\Rightarrow \frac{a}{b}=\frac{\cos{\left(\frac{\alpha+\beta}{2}\right)}}{\sin{\left(\frac{\alpha+\beta}{2}\right)}}\)
\(\Rightarrow \frac{a^2}{b^2}=\frac{\cos^2{\left(\frac{\alpha+\beta}{2}\right)}}{\sin^2{\left(\frac{\alpha+\beta}{2}\right)}}\) ➜Note উভয় পার্শে বর্গ করে।

\(\Rightarrow \frac{a^2-b^2}{a^2+b^2}=\frac{\cos^2{\left(\frac{\alpha+\beta}{2}\right)}-\sin^2{\left(\frac{\alpha+\beta}{2}\right)}}{\cos^2{\left(\frac{\alpha+\beta}{2}\right)}+\sin^2{\left(\frac{\alpha+\beta}{2}\right)}}\) ➜Note বিয়োজন-যোজন করে।

\(\Rightarrow \frac{a^2-b^2}{a^2+b^2}=\frac{\cos^2{\left(\frac{\alpha+\beta}{2}\right)}-\sin^2{\left(\frac{\alpha+\beta}{2}\right)}}{\sin^2{\left(\frac{\alpha+\beta}{2}\right)}+\cos^2{\left(\frac{\alpha+\beta}{2}\right)}}\)
\(\Rightarrow \frac{a^2-b^2}{a^2+b^2}=\frac{\cos^2{\left(\frac{\alpha+\beta}{2}\right)}-\sin^2{\left(\frac{\alpha+\beta}{2}\right)}}{1}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)

\(\Rightarrow \frac{a^2-b^2}{a^2+b^2}=\cos^2{\left(\frac{\alpha+\beta}{2}\right)}-\sin^2{\left(\frac{\alpha+\beta}{2}\right)}\)
\(\Rightarrow \cos^2{\left(\frac{\alpha+\beta}{2}\right)}-\sin^2{\left(\frac{\alpha+\beta}{2}\right)}=\frac{a^2-b^2}{a^2+b^2}\)
\(\therefore \cos{(\alpha+\beta)}=\frac{a^2-b^2}{a^2+b^2}\) ➜Note \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)

(প্রমাণিত)

দেওয়া আছে,
\(\sin{\theta}+\sin{\phi}=a\) এবং \(\cos{\theta}+\cos{\phi}=b\)
ধরি,
\(\sin{\theta}+\sin{\phi}=a .....(1)\)
এবং \(\cos{\theta}+\cos{\phi}=b .....(2)\)
\((1)\) ও \((2)\) বর্গ করে যোগ করি,
\((\sin{\theta}+\sin{\phi})^2+(\cos{\theta}+\cos{\phi})^2=a^2+b^2\)
\(\Rightarrow \sin^2{\theta}+\sin^2{\phi}+2\sin{\theta}\sin{\phi}+\cos^2{\theta}+\cos^2{\phi}+2\cos{\theta}\cos{\phi}=a^2+b^2\) ➜Note \(\because (p+q)^2=p^2+q^2+2pq\)

\(\Rightarrow (\sin^2{\theta}+\cos^2{\theta})+(\sin^2{\phi}+\cos^2{\phi})+2(\cos{\theta}\cos{\phi}+\sin{\theta}\sin{\phi})=a^2+b^2\)
\(\Rightarrow 1+1+2\cos{(\alpha-\beta)}=a^2+b^2\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(\cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)

\(\Rightarrow 2+2\cos{(\theta-\phi)}=a^2+b^2\)
\(\Rightarrow 2\{1+\cos{(\theta-\phi)}\}=a^2+b^2\)
\(\Rightarrow 2\{2\cos^2{\left(\frac{\theta-\phi}{2}\right)}\}=a^2+b^2\) ➜Note \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)

\(\Rightarrow 4\cos^2{\left(\frac{\theta-\phi}{2}\right)}=a^2+b^2\)
\(\Rightarrow \cos^2{\left(\frac{\theta-\phi}{2}\right)}=\frac{a^2+b^2}{4}\)
\(\Rightarrow \frac{1}{\cos^2{\left(\frac{\theta-\phi}{2}\right)}}=\frac{4}{a^2+b^2}\)
\(\Rightarrow \sec^2{\left(\frac{\theta-\phi}{2}\right)}=\frac{4}{a^2+b^2}\) ➜Note \(\because \frac{1}{\cos{A}}=\sec{A}\)

\(\Rightarrow 1+\tan^2{\left(\frac{\theta-\phi}{2}\right)}=\frac{4}{a^2+b^2}\) ➜Note \(\because \sec^2{A}=1+tan^2{A}\)

\(\Rightarrow \tan^2{\left(\frac{\theta-\phi}{2}\right)}=\frac{4}{a^2+b^2}-1\) ➜Note পক্ষান্তর করে।

\(\Rightarrow \tan^2{\left(\frac{\theta-\phi}{2}\right)}=\frac{4-a^2-b^2}{a^2+b^2}\)
\(\therefore \tan{\left(\frac{\theta-\phi}{2}\right)}=\pm\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}\)
(প্রমাণিত)

দেওয়া আছে,
\(\sin{x}+\sin{y}=a\) এবং \(\cos{x}+\cos{y}=b\)
ধরি,
\(\sin{x}+\sin{y}=a ........(1)\)
এবং \(\cos{x}+\cos{y}=b ......(2)\)
\((1)\) ও \((2)\) বর্গ করে যোগ করি,
\((\sin{x}+\sin{y})^2+(\cos{x}+\cos{y})^2=a^2+b^2\)
\(\Rightarrow \sin^2{x}+\sin^2{y}+2\sin{x}\sin{y}+\cos^2{x}+\cos^2{y}+2\cos{x}\cos{y}=a^2+b^2\) ➜Note \(\because (a+b)^2=a^2+b^2+2ab\)

\(\Rightarrow (\sin^2{x}+\cos^2{x})+(\sin^2{y}+\cos^2{y})+2(\cos{x}\cos{y}+\sin{x}\sin{y})=a^2+b^2\)
\(\Rightarrow 1+1+2\cos{(x-y)}=a^2+b^2\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(\cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)

\(\Rightarrow 2+2\cos{(x-y)}=a^2+b^2\)
\(\Rightarrow 2\{1+\cos{(x-y)}\}=a^2+b^2\)
\(\Rightarrow 2\{2\cos^2{\left(\frac{x-y}{2}\right)}\}=a^2+b^2\) ➜Note \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)

\(\Rightarrow 4\cos^2{\left(\frac{x-y}{2}\right)}=a^2+b^2\)
\(\Rightarrow 4\left\{1-\sin^2{\left(\frac{x-y}{2}\right)}\right\}=a^2+b^2\) ➜Note \(\because \cos^2{A}=1-\sin^2{A}\)

\(\Rightarrow 4-4\sin^2{\left(\frac{x-y}{2}\right)}=a^2+b^2\)
\(\Rightarrow 4-a^2-b^2=4\sin^2{\left(\frac{x-y}{2}\right)}\) ➜Note পক্ষান্তর করে।

\(\Rightarrow 4\sin^2{\left(\frac{x-y}{2}\right)}=4-a^2-b^2\) ➜Note পক্ষান্তর করে।

\(\Rightarrow \sin^2{\left(\frac{x-y}{2}\right)}=\frac{1}{4}(4-a^2-b^2)\)
\(\Rightarrow \sin{\left(\frac{x-y}{2}\right)}=\pm\sqrt{\frac{1}{4}(4-a^2-b^2)}\)
\(\therefore \sin{\frac{1}{2}(x-y)}=\pm\frac{1}{2}\sqrt{(4-a^2-b^2)}\)
(প্রমাণিত)

দেওয়া আছে,
\(x=\sin{\frac{\pi}{18}}\)
আমরা জানি,
\(3\sin{A}-4\sin^3{A}=\sin{3A}\)
\(\Rightarrow 3\sin{\frac{\pi}{18}}-4\sin^3{\frac{\pi}{18}}=\sin{\left(3\times\frac{\pi}{18}\right)}\) ➜Note \(A=\frac{\pi}{18}\) বসিয়ে,

\(\Rightarrow 3x-4x^3=\sin{\frac{\pi}{6}}\) ➜Note \(\because x=\sin{\frac{\pi}{18}}\)

\(\Rightarrow 3x-4x^3=\frac{1}{2}\) ➜Note \(\because \sin{\frac{\pi}{6}}=\frac{1}{2}\)

\(\Rightarrow 6x-8x^3=1\)
\(\Rightarrow 6x-8x^3-1=0\)
\(\therefore 8x^3-6x+1=0\)
এখন, \(L.S=8x^4+4x^3-6x^2-2x+\frac{1}{2}\)
\(=8x^4-6x^2+x+4x^3-3x+\frac{1}{2}\)
\(=x(8x^3-6x+1)+\frac{1}{2}(8x^3-6x+1)\)
\(=x(0)+\frac{1}{2}(0)\) ➜Note \(\because 8x^3-6x+1=0\)

\(=0+0\)
\(=0\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)

দেওয়া আছে,
\(\tan{\theta}=\frac{y}{x}\)
এখন, \(L.S=x\cos{2\theta}+y\sin{2\theta}\)
\(=x\times\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}+y\times\frac{2\tan{\theta}}{1+\tan^2{\theta}}\) ➜Note \(\because \cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)
এবং \(\sin{2A}=\frac{2\tan{A}}{1+\tan^2{A}}\)

\(=x\times\frac{1-\left(\frac{y}{x}\right)^2}{1+\left(\frac{y}{x}\right)^2}+y\times\frac{2\times\frac{y}{x}}{1+\left(\frac{y}{x}\right)^2}\) ➜Note \(\because \tan{\theta}=\frac{y}{x}\)

\(=x\times\frac{1-\frac{y^2}{x^2}}{1+\frac{y^2}{x^2}}+\frac{\frac{2y^2}{x}}{1+\frac{y^2}{x^2}}\)
\(=\frac{x-\frac{y^2}{x}}{1+\frac{y^2}{x^2}}+\frac{\frac{2y^2}{x}}{1+\frac{y^2}{x^2}}\)
\(=\frac{x^3-xy^2}{x^2+y^2}+\frac{2xy^2}{x^2+y^2}\) ➜Note উভয় ভগ্নাংশের লব ও হরকে \(x^2\) দ্বারা গুণ করে।

\(=\frac{x^3-xy^2+2xy^2}{x^2+y^2}\)
\(=\frac{x^3+xy^2}{x^2+y^2}\)
\(=\frac{x(x^2+y^2)}{x^2+y^2}\)
\(=x\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)

\(R.S=(1+\sec{2\theta})(1+\sec{2^2\theta})(1+\sec{2^3\theta})...(1+\sec{2^n\theta})\)
\(= \frac{1}{\tan{\theta}}\times\tan{\theta}\left(1+\frac{1}{\cos{2\theta}}\right)(1+\sec{2^2\theta})(1+\sec{2^3\theta})...(1+\sec{2^n\theta})\) ➜Note \(\because \sec{A}=\frac{1}{\cos{A}}\)

\(= \frac{1}{\tan{\theta}}\times\tan{\theta}\left(1+\frac{1}{\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}}\right)(1+\sec{2^2\theta})(1+\sec{2^3\theta})...(1+\sec{2^n\theta})\) ➜Note \(\because \cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(= \frac{1}{\tan{\theta}}\times\tan{\theta}\left(1+\frac{1+\tan^2{\theta}}{1-\tan^2{\theta}}\right)(1+\sec{2^2\theta})(1+\sec{2^3\theta})...(1+\sec{2^n\theta})\)
\(= \frac{1}{\tan{\theta}}\times\tan{\theta}\left(\frac{1+\tan^2{\theta}+1-\tan^2{\theta}}{1-\tan^2{\theta}}\right)(1+\sec{2^2\theta})(1+\sec{2^3\theta})...(1+\sec{2^n\theta})\)
\(= \frac{1}{\tan{\theta}}\times\tan{\theta}\left(\frac{2}{1-\tan^2{\theta}}\right)\left(1+\frac{1}{\cos{2^2\theta}}\right)(1+\sec{2^3\theta})...(1+\sec{2^n\theta})\) ➜Note \(\because \sec{A}=\frac{1}{\cos{A}}\)

\(= \frac{1}{\tan{\theta}}\times\left(\frac{2\tan{\theta}}{1-\tan^2{\theta}}\right)\left(1+\frac{1}{\frac{1-\tan^2{2\theta}}{1+\tan^2{2\theta}}}\right)(1+\sec{2^3\theta})...(1+\sec{2^n\theta})\) ➜Note \(\because\cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(= \frac{1}{\tan{\theta}}\times\tan{2\theta}\left(1+\frac{1+\tan^2{2\theta}}{1-\tan^2{2\theta}}\right)(1+\sec{2^3\theta})...(1+\sec{2^n\theta})\) ➜Note \(\because \frac{2\tan{A}}{1-\tan^2{A}}=\tan{2A}\)

\(= \frac{1}{\tan{\theta}}\times\tan{2\theta}\left(\frac{1+\tan^2{2\theta}+1-\tan^2{2\theta}}{1-\tan^2{2\theta}}\right)(1+\sec{2^3\theta})...(1+\sec{2^n\theta})\)
\(=\frac{1}{\tan{\theta}}\times\tan{2\theta}\left(\frac{2}{1-\tan^2{2\theta}}\right)\left(1+\frac{1}{\cos{2^3\theta}}\right)...(1+\sec{2^n\theta})\)➜Note \(\because \sec{A}=\frac{1}{\cos{A}}\)

\(=\frac{1}{\tan{\theta}}\times\left(\frac{2\tan{2\theta}}{1-\tan^2{2\theta}}\right)\left(1+\frac{1}{\frac{1-\tan^2{2^2\theta}}{1+\tan^2{2^2\theta}}}\right)...(1+\sec{2^n\theta})\) ➜Note \(\because \cos{2A}=\frac{1-\tan^2{A}}{1+\tan^2{A}}\)

\(=\frac{1}{\tan{\theta}}\times\tan{2^2\theta}\left(1+\frac{1+\tan^2{2^2\theta}}{1-\tan^2{2^2\theta}}\right)...(1+\sec{2^n\theta})\) ➜Note \(\because \frac{2\tan{A}}{1-\tan^2{A}}=\tan{2A}\)

\(=\frac{1}{\tan{\theta}}\times\tan{2^2\theta}\left(\frac{1+\tan^2{2^2\theta}+1-\tan^2{2^2\theta}}{1-\tan^2{2^2\theta}}\right)...(1+\sec{2^n\theta})\)
\(=\frac{1}{\tan{\theta}}\times\tan{2^2\theta}\left(\frac{2}{1-\tan^2{2^2\theta}}\right)...(1+\sec{2^n\theta})\)
\(=\frac{1}{\tan{\theta}}\times\left(\frac{2\tan{2^2\theta}}{1-\tan^2{2^2\theta}}\right)...(1+\sec{2^n\theta})\)
\(=\frac{1}{\tan{\theta}}\times\tan{2^3\theta}...(1+\sec{2^n\theta})\) ➜Note \(\because \frac{2\tan{A}}{1-\tan^2{A}}=\tan{2A}\)

\(.........................\)
\(.........................\)
\(=\frac{1}{\tan{\theta}}\times\tan{2^n\theta}\)
\(=\frac{\tan{2^n\theta}}{\tan{\theta}}\)
\(=L.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)

দেওয়া আছে,
\(13\theta=\pi\)
\(L.S=\cos{\theta}\cos{2\theta}\cos{3\theta}\cos{4\theta}\cos{5\theta}\cos{6\theta}\)
\(=\frac{1}{2\sin{\theta}}(2\sin{\theta}\cos{\theta})\cos{2\theta}\cos{3\theta}\cos{4\theta}\cos{5\theta}\cos{6\theta}\)
\(=\frac{1}{2\sin{\theta}}\sin{2\theta}\cos{2\theta}\cos{3\theta}\cos{4\theta}\cos{5\theta}\cos{6\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{1}{2^2\sin{\theta}}(2\sin{2\theta}\cos{2\theta})\cos{3\theta}\cos{4\theta}\cos{5\theta}\cos{6\theta}\)
\(=\frac{1}{2^2\sin{\theta}}\sin{4\theta}\cos{3\theta}\cos{4\theta}\cos{5\theta}\cos{6\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{1}{2^3\sin{\theta}}(2\sin{4\theta}\cos{4\theta})\cos{3\theta}\cos{5\theta}\cos{6\theta}\)
\(=\frac{1}{2^3\sin{\theta}}\sin{8\theta}\cos{3\theta}\cos{5\theta}\cos{6\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{1}{2^3\sin{\theta}}\sin{(13\theta-5\theta)}\cos{3\theta}\cos{5\theta}\cos{6\theta}\)
\(=\frac{1}{2^3\sin{\theta}}\sin{(\pi-5\theta)}\cos{3\theta}\cos{5\theta}\cos{6\theta}\) ➜Note \(\because 13\theta=\pi\)

\(=\frac{1}{2^3\sin{\theta}}\sin{\left(\frac{\pi}{2}\times2-5\theta\right)}\cos{3\theta}\cos{5\theta}\cos{6\theta}\) ➜Note \(\because \pi-5\theta=\frac{\pi}{2}\times2-5\theta\)

\(=\frac{1}{2^3\sin{\theta}}\sin{5\theta}\cos{3\theta}\cos{5\theta}\cos{6\theta}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=\frac{1}{2^4\sin{\theta}}(2\sin{5\theta}\cos{5\theta})\cos{3\theta}\cos{6\theta}\)
\(=\frac{1}{2^4\sin{\theta}}\sin{10\theta}\cos{3\theta}\cos{6\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{1}{2^4\sin{\theta}}\sin{(13\theta-3\theta)}\cos{3\theta}\cos{6\theta}\)
\(=\frac{1}{2^4\sin{\theta}}\sin{(\pi-3\theta)}\cos{3\theta}\cos{6\theta}\) ➜Note \(\because 13\theta=\pi\)

\(=\frac{1}{2^4\sin{\theta}}\sin{\left(\frac{\pi}{2}\times2-3\theta\right)}\cos{3\theta}\cos{6\theta}\) ➜Note \(\because \pi-3\theta=\frac{\pi}{2}\times2-3\theta\)

\(=\frac{1}{2^4\sin{\theta}}\sin{3\theta}\cos{3\theta}\cos{6\theta}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=\frac{1}{2^5\sin{\theta}}(2\sin{3\theta}\cos{3\theta})\cos{6\theta}\)
\(=\frac{1}{2^5\sin{\theta}}\sin{6\theta}\cos{6\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{1}{2^6\sin{\theta}}(2\sin{6\theta}\cos{6\theta})\)
\(=\frac{1}{2^6\sin{\theta}}\sin{12\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{1}{2^6\sin{\theta}}\sin{(13\theta-\theta)}\)
\(=\frac{1}{2^6\sin{\theta}}\sin{(\pi-\theta)}\) ➜Note \(\because 13\theta=\pi\)

\(=\frac{1}{2^6\sin{\theta}}\sin{\left(\frac{\pi}{2}\times2-\theta\right)}\) ➜Note \(\because \pi-\theta=\frac{\pi}{2}\times2-\theta\)

\(=\frac{\sin{\theta}}{2^6\sin{\theta}}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=\frac{1}{2^6}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)

দেওয়া আছে,
\(\theta=\frac{\pi}{2^n+1}\)
\(\Rightarrow 2^n\theta+\theta=\pi\)
\(\therefore 2^n\theta=\pi-\theta\)
\(L.S=2^n\cos{\theta}\cos{2\theta}\cos{2^2\theta}....\cos{2^{n-1}\theta}\)
\(=\frac{2^{n-1}}{\sin{\theta}}(2\sin{\theta}\cos{\theta})\cos{2\theta}\cos{2^2\theta}....\cos{2^{n-1}\theta}\)
\(=\frac{2^{n-1}}{\sin{\theta}}\sin{2\theta}\cos{2\theta}\cos{2^2\theta}....\cos{2^{n-1}\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{2^{n-2}}{\sin{\theta}}(2\sin{2\theta}\cos{2\theta})\cos{2^2\theta}....\cos{2^{n-1}\theta}\)
\(=\frac{2^{n-2}}{\sin{\theta}}\sin{2^2\theta}\cos{2^2\theta}....\cos{2^{n-1}\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(=\frac{2^{n-3}}{\sin{\theta}}(2\sin{2^2\theta}\cos{2^2\theta})....\cos{2^{n-1}\theta}\)
\(=\frac{2^{n-3}}{\sin{\theta}}\sin{2^3\theta}....\cos{2^{n-1}\theta}\) ➜Note \(\because 2\sin{A}\cos{A}=\sin{2A}\)

\(.................................\)
\(.................................\)
\(=\frac{2^{n-n}}{\sin{\theta}}\sin{2^n\theta}\)
\(=\frac{2^{0}}{\sin{\theta}}\sin{(\pi-\theta)}\) ➜Note \(\because 2^n\theta=\pi-\theta\)

\(=\frac{1}{\sin{\theta}}\sin{\left(\frac{\pi}{2}\times2-\theta\right)}\) ➜Note \(\because \pi-\theta=\frac{\pi}{2}\times2-\theta\)
এবং \(2^{0}=1\)

\(=\frac{\sin{\theta}}{\sin{\theta}}\) ➜Note
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(2\) একটি জোড় সংখ্যা,
তাই অনুপাতের পরিবর্তন হয়নি।

\(=1\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)

দেওয়া আছে,
\((A+B)\ne{0}\) এবং \(\sin{A}+\sin{B}=2\sin{(A+B)}\)
\(\Rightarrow \sin{A}+\sin{B}=2\sin{(A+B)}\)
\(\Rightarrow \frac{\sin{A}+\sin{B}}{\sin{(A+B)}}=2\) ➜Note পক্ষান্তর করে।

\(\Rightarrow \frac{\sin{A}+\sin{B}-\sin{(A+B)}}{\sin{A}+\sin{B}+\sin{(A+B)}}=\frac{2-1}{2+1}\) ➜Note বিয়োজন-যোজন করে।

\(\Rightarrow \frac{2\sin{\frac{A+B}{2}}\cos{\frac{A-B}{2}}-2\sin{\frac{A+B}{2}}\cos{\frac{A+B}{2}}}{2\sin{\frac{A+B}{2}}\cos{\frac{A-B}{2}}+2\sin{\frac{A+B}{2}}\cos{\frac{A+B}{2}}}=\frac{1}{3}\) ➜Note \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
এবং \(\sin{2A}=2\sin{A}\cos{A}\)

\(\Rightarrow \frac{2\sin{\frac{A+B}{2}}\left(\cos{\frac{A-B}{2}}-\cos{\frac{A+B}{2}}\right)}{2\sin{\frac{A+B}{2}}\left(\cos{\frac{A-B}{2}}+\cos{\frac{A+B}{2}}\right)}=\frac{1}{3}\)
\(\Rightarrow \frac{\cos{\frac{A-B}{2}}-\cos{\frac{A+B}{2}}}{\cos{\frac{A-B}{2}}+\cos{\frac{A+B}{2}}}=\frac{1}{3}\)
\(\Rightarrow \frac{\cos{\left(\frac{A}{2}-\frac{B}{2}\right)}-\cos{\left(\frac{A}{2}+\frac{B}{2}\right)}}{\cos{\left(\frac{A}{2}-\frac{B}{2}\right)}+\cos{\left(\frac{A}{2}+\frac{B}{2}\right)}}=\frac{1}{3}\)
\(\Rightarrow \frac{2\sin{\frac{A}{2}}\sin{\frac{B}{2}}}{2\cos{\frac{A}{2}}\cos{\frac{B}{2}}}=\frac{1}{3}\) ➜Note \(\because \cos{(A-B)}-\cos{(A+B)}=2\sin{A}\sin{B}\)
এবং \(\cos{(A-B)}+\cos{(A+B)}=2\cos{A}\cos{B}\)

\(\Rightarrow \frac{\sin{\frac{A}{2}}\sin{\frac{B}{2}}}{\cos{\frac{A}{2}}\cos{\frac{B}{2}}}=\frac{1}{3}\)
\(\therefore \tan{\frac{A}{2}}\tan{\frac{B}{2}}=\frac{1}{3}\) ➜Note \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)

(দেখানো হলো)

দেওয়া আছে,
\(\sin{x}+\sin{y}=a\) এবং \(\cos{x}+\cos{y}=b\)
ধরি,
\(\sin{x}+\sin{y}=a ........(1)\)
এবং \(\cos{x}+\cos{y}=b ......(2)\)
\((1)\) ও \((2)\) বর্গ করে যোগ করি,
\((\sin{x}+\sin{y})^2+(\cos{x}+\cos{y})^2=a^2+b^2\)
\(\Rightarrow \sin^2{x}+\sin^2{y}+2\sin{x}\sin{y}+\cos^2{x}+\cos^2{y}+2\cos{x}\cos{y}=a^2+b^2\) ➜Note \(\because (a+b)^2=a^2+b^2+2ab\)

\(\Rightarrow (\sin^2{x}+\cos^2{x})+(\sin^2{y}+\cos^2{y})+2(\cos{x}\cos{y}+\sin{x}\sin{y})=a^2+b^2\)
\(\Rightarrow 1+1+2\cos{(\alpha-\beta)}=a^2+b^2\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(\cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)

\(\Rightarrow 2+2\cos{(x-y)}=a^2+b^2\)
\(\Rightarrow 2\{1+\cos{(x-y)}\}=a^2+b^2\)
\(\Rightarrow 2\{2\cos^2{\left(\frac{x-y}{2}\right)}\}=a^2+b^2\) ➜Note \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)

\(\Rightarrow 4\cos^2{\left(\frac{x-y}{2}\right)}=a^2+b^2\)
\(\Rightarrow \cos^2{\left(\frac{x-y}{2}\right)}=\frac{1}{4}(a^2+b^2)\)
\(\Rightarrow \cos{\left(\frac{x-y}{2}\right)}=\pm\sqrt{\frac{1}{4}(a^2+b^2)}\)
\(\therefore \cos{\frac{1}{2}(x-y)}=\pm\frac{1}{2}\sqrt{a^2+b^2}\)
(প্রমাণিত)

দেওয়া আছে,
\(a\sin{x}+b\sin{y}=c=a\cos{x}+b\cos{y}\)
ধরি,
\(a\sin{x}+b\sin{y}=c ........(1)\)
এবং \(a\cos{x}+b\cos{y}=c ......(2)\)
\((1)\) ও \((2)\) বর্গ করে যোগ করি,
\((a\sin{x}+b\sin{y})^2+(a\cos{x}+b\cos{y})^2=c^2+c^2\)
\(\Rightarrow a^2\sin^2{x}+b^2\sin^2{y}+2ab\sin{x}\sin{y}+a^2\cos^2{x}+b^2\cos^2{y}+2ab\cos{x}\cos{y}=2c^2\) ➜Note \(\because (a+b)^2=a^2+b^2+2ab\)

\(\Rightarrow a^2(\sin^2{x}+\cos^2{x})+b^2(\sin^2{y}+\cos^2{y})+2ab(\cos{x}\cos{y}+\sin{x}\sin{y})=2c^2\)
\(\Rightarrow a^2.1+b^2.1+2ab\cos{(x-y)}=2c^2\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(\cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)

\(\Rightarrow a^2+b^2+2ab\cos{(x-y)}=2c^2\)
\(\Rightarrow 2ab\cos{(x-y)}=2c^2-(a^2+b^2)\)
\(\Rightarrow \cos{(x-y)}=\frac{2c^2-(a^2+b^2)}{2ab}\)
\(\Rightarrow 2\cos^2{\frac{x-y}{2}}-1=\frac{2c^2-(a^2+b^2)}{2ab}\) ➜Note \(\because \cos{2A}=2\cos^2{A}-1\)

\(\Rightarrow 2\cos^2{\frac{x-y}{2}}=\frac{2c^2-(a^2+b^2)}{2ab}+1\)
\(\Rightarrow 2\cos^2{\frac{x-y}{2}}=\frac{2c^2-(a^2+b^2)+2ab}{2ab}\)
\(\Rightarrow \cos^2{\frac{x-y}{2}}=\frac{2c^2-(a^2+b^2-2ab)}{4ab}\)
\(\Rightarrow \cos^2{\frac{x-y}{2}}=\frac{2c^2-(a-b)^2}{4ab}\) ➜Note \(\because a^2+b^2-2ab=(a-b)^2\)

\(\Rightarrow \cos{\frac{x-y}{2}}=\pm\sqrt{\frac{2c^2-(a-b)^2}{4ab}}\)
\(\therefore \cos{\frac{1}{2}(x-y)}=\pm\sqrt{\frac{2c^2-(a-b)^2}{4ab}}\)
(দেখানো হলো)