প্রমাণঃ
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 \(\because a^2+b^2-2ab=(a-b)^2\)

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

প্রমাণঃ
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 \(\because \sec^2{A}=1+\tan^2{A}\)

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

প্রমাণঃ
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।

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

প্রমাণঃ
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।
\(\frac{1-\cos{2A}}{1+\cos{2A}}=\frac{2\sin^2{A}}{2\cos^2{A}}\)
\(=\frac{\sin^2{A}}{\cos^2{A}}\)
\(=\tan^2{A}\)
\(\therefore \frac{1-\cos{2A}}{1+\cos{2A}}=\tan^2{A}\)
\(\frac{1-\cos{2A}}{1+\cos{2A}}=\tan^2{A}\)

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

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

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

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

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

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

প্রমাণঃ
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,
\(\tan{(A+A+A)}=\frac{\tan{A}+\tan{A}+\tan{A}-\tan{A}\tan{A}\tan{A}}{1-\tan{A}\tan{A}-\tan{A}\tan{A}-\tan{A}\tan{A}}\)
\(\Rightarrow \tan{3A}=\frac{3\tan{A}-\tan^3{A}}{1-\tan^2{A}-\tan^2{A}-\tan^2{A}}\)
\(\therefore \tan{3A}=\frac{3\tan{A}-\tan^3{A}}{1-3\tan^2{A}}\)
\(\tan{3A}=\frac{3\tan{A}-\tan^3{A}}{1-3\tan^2{A}}\)

প্রমাণঃ
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,
\(\cot{(A+A+A)}=\frac{\cot{A}\cot{A}\cot{A}-\cot{A}-\cot{A}-\cot{A}}{\cot{A}\cot{A}+\cot{A}\cot{A}+\cot{A}\cot{A}-1}\)
\(\Rightarrow \cot{3A}=\frac{\cot^3{A}-3\cot{A}}{\cot^2{A}+\cot^2{A}+\cot^2{A}-1}\)
\(\therefore \cot{3A}=\frac{\cot^3{A}-3\cot{A}}{3\cot^2{A}-1}\)
\(\cot{3A}=\frac{\cot^3{A}-3\cot{A}}{3\cot^2{A}-1}\)

প্রমাণঃ
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 \(\because \frac{\cos{A}}{\sin{A}}=\cot{A}\)

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