প্রমাণঃ
মনে করি, কোনো ঘূর্ণায়মান রশ্মি আদি অবস্থান \(OX\) হতে \(O\) বিন্দুর সাপেক্ষে ঘড়ির কাঁটার ঘূর্ণনের বিপরীত দিকে ঘুরে \(\angle{XOY}=A, \ (A\lt90^{o})\) এবং একই দিকে আরো ঘুরে
\(\angle{YOZ}=B, \ (B\lt90^{o})\) কোণ উৎপন্ন করে।
\(\therefore \angle{XOZ}=\angle{XOY}+\angle{YOZ}\)
\(=A+B\)
ঘূর্ণায়মান রেখার শেষ অবস্থান \(OZ\) এর উপর \(P\) যে কোনো বিন্দু নিয়ে \(OX\) এবং \(OY\) রেখার উপর যথাক্রমে \(PQ\) এবং \(PR\) লম্ব অংকন করি। আবার \(R\) বিন্দু হতে \(OX\) এবং \(PQ\) রেখার উপর যথাক্রমে \(RS\) এবং \(RT\) লম্ব অংকন করি।
তাহলে \(QSRT\) একটি আয়তক্ষেত্র,
যার \(RT=QS\) ও \(RS=TQ\)
আবার, \(TR||OX\)
\(OR\) উহাদের ছেদক।
\(\therefore \angle{TRO}=\text{একান্তর }\angle{XOY}=A\)
এখন \(\triangle{PTR}\) ত্রিভুজে
\(\angle{TPR}=90^{o}-\angle{PRT} \) ➜Note \(\because \angle{PRO}=90^{o}\)

\(=\angle{TRO} \)
\(=A\)
\(\therefore \angle{TPR}=A\)
\(\triangle{ROS} \)-এ
\(\sin{A}=\frac{RS}{OR}, \ \cos{A}=\frac{OS}{OR}\)
\(\triangle{POR} \)-এ
\(\sin{B}=\frac{PR}{OP}, \ \cos{B}=\frac{OR}{OP}\)
এবং \(\triangle{PTR}\)-এ
\(\sin{A}=\frac{TR}{PR}, \ \cos{A}=\frac{PT}{PR}\)
\(POQ \) সমকোণী ত্রিভুজ থেকে,
\(\angle{POQ}=A+B\)
\(\sin{(A+B)}=\frac{PQ}{OP}\)
\(=\frac{QT+TP}{OP}\)
\(=\frac{RS+TP}{OP}\) ➜Note \(\because QT=RS\)

\(=\frac{RS}{OP}+\frac{TP}{OP}\)
\(=\frac{RS}{OR}\times\frac{OR}{OP}+\frac{TP}{PR}\times\frac{PR}{OP}\)
\(=\sin{A}\cos{B}+\cos{A}\sin{B}\) ➜Note \(\because \sin{A}=\frac{RS}{OR}\)
\(\cos{B}=\frac{OR}{OP}\)
\(\cos{A}=\frac{PT}{PR}\)
এবং \(\sin{B}=\frac{PR}{OP}\)

\(\therefore \sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
\(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
আবার, \(\cos{(A+B)}=\frac{OQ}{OP}\)
\(=\frac{OS-QS}{OP}\)
\(=\frac{OS-TR}{OP}\) ➜Note \(\because QS=TR\)

\(=\frac{OS}{OP}-\frac{TR}{OP}\)
\(=\frac{OS}{OR}\times\frac{OR}{OP}-\frac{TR}{PR}\times\frac{PR}{OP}\)
\(=\cos{A}\cos{B}-\sin{A}\sin{B}\) ➜Note \(\because \cos{A}=\frac{PT}{PR}\)
\(\cos{B}=\frac{OR}{OP}\)
\(\sin{A}=\frac{RS}{OR}\)
এবং \(\sin{B}=\frac{PR}{OP}\)

\(\therefore \cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
\(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
মনে করি, \(PQRS\) আয়তক্ষেত্রে \(TOQ\) সমকোণী ত্রিভুজ অবস্থিত,
যেখানে \(\angle{TOQ}=90^{o}\)
\(\angle{OQT}=A, \ \angle{OQP}=B\)
এবং \(QT=1\) একক।
\(PQ=SR\) ও \(PS=QR\)
তাহলে, \(\angle{POQ}=90^{o}-B\)
\(\angle{SOT}=90^{o}-\angle{POQ}\)
\(=90^{o}-(90^{o}-B)\)
\(=90^{o}-90^{o}+B\)
\(=B\)
\(\therefore \angle{SOT}=B\)
\(\angle{TQR}=90^{o}-(A+B)\)
\(\angle{QTR}=90^{o}-\angle{TQR}\)
\(=90^{o}-\{90^{o}-(A+B)\}\)
\(=90^{o}-90^{o}+(A+B)\)
\(\therefore \angle{QTR}=A+B\)
এখন \(\sin{A}=\frac{OT}{QT}\)
\(\Rightarrow \sin{A}=\frac{OT}{1}\) ➜Note \(\because QT=1\)

\(\therefore OT=\sin{A}\)
আবার, \(\cos{A}=\frac{OQ}{QT}\)
\(\Rightarrow \cos{A}=\frac{OQ}{1}\) ➜Note \(\because QT=1\)

\(\therefore OQ=\cos{A}\)
\(OPQ\) সমকোণী ত্রিভুজ হতে,
\(\sin{B}=\frac{OP}{OQ}\)
\(\Rightarrow OP=OQ\sin{B}\)
\(\therefore OP=\cos{A}\sin{B}\) ➜Note \(\because OQ=\cos{A}\)

আবার, \(\cos{B}=\frac{PQ}{OQ}\)
\(\Rightarrow PQ=OQ\cos{B}\)
\(\therefore PQ=\cos{A}\cos{B}\) ➜Note \(\because OQ=\cos{A}\)

অনুরূপভাবে, \(OST\) সমকোণী ত্রিভুজ হতে,
\(OS=\sin{A}\cos{B}\)
এবং \(ST=\sin{A}\sin{B}\)
এখন \(TQR\) সমকোণী ত্রিভুজ হতে,
\(\sin{(A+B)}=\sin{QTR}\)
\(=\frac{QR}{QT}\)
\(=\frac{PS}{1}\) ➜Note \(\because QR=PS\)
এবং \(QT=1\)

\(=PS\)
\(=OS+OP\)
\(=\sin{A}\cos{B}+\cos{A}\sin{B}\) ➜Note \(\because OS=\sin{A}\cos{B}\)
এবং \(OP=\cos{A}\sin{B}\)

\(\therefore \sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
\(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
আবার, \(\cos{(A+B)}=\cos{QTR}\)
\(=\frac{TR}{QT}\)
\(=\frac{TR}{1}\) ➜Note \(\because QT=1\)

\(=TR\)
\(=RS-ST\)
\(=PQ-ST\) ➜Note \(\because RS=PQ\)

\(=\cos{A}\cos{B}-\sin{A}\sin{B}\) ➜Note \(\because PQ=\cos{A}\cos{B}\)
এবং \(ST=\sin{A}\sin{B}\)

\(\therefore \cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
\(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)

প্রমাণঃ
মনে করি, কোনো ঘূর্ণায়মান রশ্মি আদি অবস্থান \(OX\) হতে \(O\) বিন্দুর সাপেক্ষে ঘড়ির কাঁটার ঘূর্ণনের বিপরীত দিকে ঘুরে \(\angle{XOY}=A, \ (A\lt90^{o})\) এবং \(OY\) অবস্থান থেকে ঘড়ির কাঁটার ঘূর্ণনের দিকে ঘুরে \(\angle{YOZ}=B, \ (B\lt90^{o})\) কোণ উৎপন্ন করে।
\(\therefore \angle{XOZ}=\angle{XOY}-\angle{YOZ}\)
\(=A-B\)
ঘূর্ণায়মান রেখার শেষ অবস্থান \(OZ\) এর উপর \(P\) যে কোনো বিন্দু নিয়ে \(OX\) এবং \(OY\) রেখার উপর যথাক্রমে \(PQ\) এবং \(PR\) লম্ব অংকন করি। আবার \(R\) বিন্দু হতে \(OX\) এবং \(PQ\) রেখার বর্ধিতাংশের উপর যথাক্রমে \(RS\) এবং \(RT\) লম্ব অংকন করি।
তাহলে \(QSRT\) একটি আয়তক্ষেত্র,
যার \(RT=QS\) ও \(RS=TQ\)
আবার, \(TR||OX\)
\(OY\) উহাদের ছেদক।
\(\therefore \angle{YRT}=\text{অনুরূপ }\angle{XOY}=A\)
এখন \(\triangle{PTR}\) ত্রিভুজে
\(\angle{TPR}=90^{o}-\angle{PRT} \) ➜Note \(\because \angle{PRY}=90^{o}\)

\(=\angle{YRT} \)
\(=A\)
\(\therefore \angle{TPR}=A\)
\(\triangle{ROS} \)-এ
\(\sin{A}=\frac{RS}{OR}, \ \cos{A}=\frac{OS}{OR}\)
\(\triangle{POR} \)-এ
\(\sin{B}=\frac{PR}{OP}, \ \cos{B}=\frac{OR}{OP}\)
এবং \(\triangle{PTR}\)-এ
\(\sin{A}=\frac{TR}{PR}, \ \cos{A}=\frac{PT}{PR}\)
\(POQ \) সমকোণী ত্রিভুজ থেকে,
\(\angle{POQ}=A-B\)
\(\sin{(A-B)}=\frac{PQ}{OP}\)
\(=\frac{TQ-PT}{OP}\)
\(=\frac{RS-PT}{OP}\) ➜Note \(\because TQ=RS\)

\(=\frac{RS}{OP}-\frac{PT}{OP}\)
\(=\frac{RS}{OR}\times\frac{OR}{OP}-\frac{PT}{PR}\times\frac{PR}{OP}\)
\(=\sin{A}\cos{B}-\cos{A}\sin{B}\) ➜Note \(\because \sin{A}=\frac{RS}{OR}\)
\(\cos{B}=\frac{OR}{OP}\)
\(\cos{A}=\frac{PT}{PR}\)
এবং \(\sin{B}=\frac{PR}{OP}\)

\(\therefore \sin{(A-B)}=\sin{A}\cos{B}-\cos{A}\sin{B}\)
\(\sin{(A-B)}=\sin{A}\cos{B}-\cos{A}\sin{B}\)
আবার, \(\cos{(A-B)}=\frac{OQ}{OP}\)
\(=\frac{OS+QS}{OP}\)
\(=\frac{OS+TR}{OP}\) ➜Note \(\because QS=TR\)

\(=\frac{OS}{OP}+\frac{TR}{OP}\)
\(=\frac{OS}{OR}\times\frac{OR}{OP}+\frac{TR}{PR}\times\frac{PR}{OP}\)
\(=\cos{A}\cos{B}+\sin{A}\sin{B}\) ➜Note \(\because \cos{A}=\frac{PT}{PR}\)
\(\cos{B}=\frac{OR}{OP}\)
\(\sin{A}=\frac{RS}{OR}\)
এবং \(\sin{B}=\frac{PR}{OP}\)

\(\therefore \cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)
\(\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)
মনে করি, \(PQRS\) আয়তক্ষেত্রে \(NOQ\) সমকোণী ত্রিভুজ অবস্থিত,
যেখানে \(\angle{NOQ}=90^{o}\)
\(\angle{OQN}=B, \ \angle{OQR}=A\)
\(\angle{POQ}=\text{একান্তর }\angle{OQR}=A\)
\(\therefore \angle{POQ}=A\)
\(\angle{ONS}=90^{o}-(90^{o}-A)\)
\(=90^{o}-90^{o}+A\)
\(\therefore \angle{ONS}=A\)
এবং \(QN=1\) একক।
\(PQ=SR\) ও \(PS=QR\)
\(\therefore \angle{NQR}=A-B\)
এখন \(\sin{B}=\frac{ON}{QN}\)
\(\Rightarrow \sin{A}=\frac{ON}{1}\) ➜Note \(\because QN=1\)

\(\therefore ON=\sin{B}\)
আবার, \(\cos{B}=\frac{OQ}{QN}\)
\(\Rightarrow \cos{B}=\frac{OQ}{1}\) ➜Note \(\because QN=1\)

\(\therefore OQ=\cos{B}\)
\(OPQ\) সমকোণী ত্রিভুজ হতে,
\(\cos{A}=\frac{OP}{OQ}\)
\(\Rightarrow OP=OQ\cos{A}\)
\(\therefore OP=\cos{A}\cos{B}\) ➜Note \(\because OQ=\cos{B}\)

আবার, \(\sin{A}=\frac{PQ}{OQ}\)
\(\Rightarrow PQ=OQ\sin{A}\)
\(\therefore PQ=\sin{A}\cos{B}\) ➜Note \(\because OQ=\cos{B}\)

\(\therefore SR=PQ=\sin{A}\cos{B}\)
অনুরূপভাবে, \(OSN\) সমকোণী ত্রিভুজ হতে,
\(OS=\sin{A}\sin{B}\)
এবং \(SN=\cos{A}\sin{B}\)
এখন \(NQR\) সমকোণী ত্রিভুজ হতে,
\(\sin{(A-B)}=\sin{NQR}\)
\(=\frac{NR}{NQ}\)
\(=\frac{NR}{1}\) ➜Note \(\because NQ=1\)

\(=NR\)
\(=SR-SN\)
\(=\sin{A}\cos{B}-\cos{A}\sin{B}\) ➜Note \(\because SR=\sin{A}\cos{B}\)
এবং \(SN=\cos{A}\sin{B}\)

\(\therefore \sin{(A-B)}=\sin{A}\cos{B}-\cos{A}\sin{B}\)
\(\sin{(A-B)}=\sin{A}\cos{B}-\cos{A}\sin{B}\)
আবার, \(\cos{(A-B)}=\cos{NQR}\)
\(=\frac{QR}{QN}\)
\(=\frac{PS}{1}\) ➜Note \(\because QN=1\)
\(QR=PS\)

\(=PS\)
\(=OP+OS\)
\(=\cos{A}\cos{B}+\sin{A}\sin{B}\) ➜Note \(\because OP=\cos{A}\cos{B}\)
এবং \(OS=\sin{A}\sin{B}\)

\(\therefore \cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)
\(\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)

প্রমাণঃ
আমরা জানি,
\(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
\(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
এখন, \(\tan{(A+B)}=\frac{\sin{(A+B)}}{\cos{(A+B)}}\)
\(=\frac{\sin{A}\cos{B}+\cos{A}\sin{B}}{\cos{A}\cos{B}-\sin{A}\sin{B}}\) ➜Note \(\because \sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
এবং \(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)

\(=\frac{\frac{\sin{A}\cos{B}}{\cos{A}\cos{B}}+\frac{\cos{A}\sin{B}}{\cos{A}\cos{B}}}{\frac{\cos{A}\cos{B}}{\cos{A}\cos{B}}-\frac{\sin{A}\sin{B}}{\cos{A}\cos{B}}}\) ➜Note লব ও হরের সহিত \(\cos{A}\cos{B}\) ভাগ করে।

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

\(\therefore \tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)
\(\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)
আবার,
আমরা জানি,
\(\sin{(A-B)}=\sin{A}\cos{B}-\cos{A}\sin{B}\)
\(\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)
এখন, \(\tan{(A-B)}=\frac{\sin{(A-B)}}{\cos{(A-B)}}\)
\(=\frac{\sin{A}\cos{B}-\cos{A}\sin{B}}{\cos{A}\cos{B}+\sin{A}\sin{B}}\) ➜Note \(\because \sin{(A-B)}=\sin{A}\cos{B}-\cos{A}\sin{B}\)
এবং \(\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)

\(=\frac{\frac{\sin{A}\cos{B}}{\cos{A}\cos{B}}-\frac{\cos{A}\sin{B}}{\cos{A}\cos{B}}}{\frac{\cos{A}\cos{B}}{\cos{A}\cos{B}}+\frac{\sin{A}\sin{B}}{\cos{A}\cos{B}}}\) ➜Note লব ও হরের সহিত \(\cos{A}\cos{B}\) ভাগ করে।

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

\(\therefore \tan{(A-B)}=\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)
\(\tan{(A-B)}=\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)

প্রমাণঃ
আমরা জানি,
\(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
\(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
এখন, \(\cot{(A+B)}=\frac{\cos{(A+B)}}{\sin{(A+B)}}\)
\(=\frac{\cos{A}\cos{B}-\sin{A}\sin{B}}{\sin{A}\cos{B}+\cos{A}\sin{B}}\) ➜Note \(\because \cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
এবং \(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)

\(=\frac{\frac{\cos{A}\cos{B}}{\sin{A}\sin{B}}-\frac{\sin{A}\sin{B}}{\sin{A}\sin{B}}}{\frac{\sin{A}\cos{B}}{\sin{A}\sin{B}}+\frac{\cos{A}\sin{B}}{\sin{A}\sin{B}}}\) ➜Note লব ও হরের সহিত \(\sin{A}\sin{B}\) ভাগ করে।

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

\(\therefore \cot{(A+B)}=\frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}}\)
\(\cot{(A+B)}=\frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}}\)
আবার,
আমরা জানি,
\(\sin{(A-B)}=\sin{A}\cos{B}-\cos{A}\sin{B}\)
\(\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)
এখন, \(\cot{(A-B)}=\frac{\cos{(A-B)}}{\sin{(A-B)}}\)
\(=\frac{\cos{A}\cos{B}+\sin{A}\sin{B}}{\sin{A}\cos{B}-\cos{A}\sin{B}}\) ➜Note \(\because \cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)
এবং \(\sin{(A-B)}=\sin{A}\cos{B}-\cos{A}\sin{B}\)

\(=\frac{\frac{\cos{A}\cos{B}}{\sin{A}\sin{B}}+\frac{\sin{A}\sin{B}}{\sin{A}\sin{B}}}{\frac{\sin{A}\cos{B}}{\sin{A}\sin{B}}-\frac{\cos{A}\sin{B}}{\sin{A}\sin{B}}}\) ➜Note লব ও হরের সহিত \(\sin{A}\sin{B}\) ভাগ করে।

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

\(\therefore \cot{(A-B)}=\frac{\cot{A}\cot{B}+1}{\cot{B}-\cot{A}}\)
\(\cot{(A-B)}=\frac{\cot{A}\cot{B}+1}{\cot{B}-\cot{A}}\)

প্রমাণঃ
আমরা জানি,
\(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
\(\sin{(A-B)}=\sin{A}\cos{B}-\cos{A}\sin{B}\)
এখন, \(\sin{(A+B)}\sin{(A-B)}=(\sin{A}\cos{B}+\cos{A}\sin{B})(\sin{A}\cos{B}-\cos{A}\sin{B})\)
\(=\sin^2{A}\cos^2{B}-\cos^2{A}\sin^2{B}\) ➜Note \(\because (a+b)(a-b)=a^2-b^2\)

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

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

\(=-\cos^2{A}+\cos^2{B}\)
\(=\cos^2{B}-\cos^2{A}\)
\(\therefore \sin{(A+B)}\sin{(A-B)}=\cos^2{B}-\cos^2{A}\)
\(\sin{(A+B)}\sin{(A-B)}=\sin^2{A}-\sin^2{B}=\cos^2{B}-\cos^2{A}\)

প্রমাণঃ
আমরা জানি,
\(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
\(\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)
এখন, \(\cos{(A+B)}\cos{(A-B)}=(\cos{A}\cos{B}-\sin{A}\sin{B})(\cos{A}\cos{B}+\sin{A}\sin{B})\)
\(=\cos^2{A}\cos^2{B}-\sin^2{A}\sin^2{B}\) ➜Note \(\because (a-b)(a+b)=a^2-b^2\)

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

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

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

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

প্রমাণঃ
আমরা জানি,
\(\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)
\(\tan{(A-B)}=\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)
এখন, \(\tan{(A+B)}\tan{(A-B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\times\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)
\(=\frac{(\tan{A}+\tan{B})(\tan{A}-\tan{B})}{(1-\tan{A}\tan{B})(1+\tan{A}\tan{B})}\)
\(=\frac{\tan^2{A}-\tan^2{B}}{1-\tan^2{A}\tan^2{B}}\) ➜Note \(\because (a-b)(a+b)=a^2-b^2\)

\(\therefore \tan{(A+B)}\tan{(A-B)}=\frac{\tan^2{A}-\tan^2{B}}{1-\tan^2{A}\tan^2{B}}\)
\(\tan{(A+B)}\tan{(A-B)}=\frac{\tan^2{A}-\tan^2{B}}{1-\tan^2{A}\tan^2{B}}\)

প্রমাণঃ
আমরা জানি,
\(\cot{(A+B)}=\frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}}\)
\(\cot{(A-B)}=\frac{\cot{A}\cot{B}+1}{\cot{B}-\cot{A}}\)
এখন, \(\cot{(A+B)}\cot{(A-B)}=\frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}}\times\frac{\cot{A}\cot{B}+1}{\cot{B}-\cot{A}}\)
\(=\frac{(\cot{A}\cot{B}-1)(\cot{A}\cot{B}+1)}{(\cot{B}+\cot{A})(\cot{B}-\cot{A})}\)
\(=\frac{\cot^2{A}\cot^2{B}-1}{\cot^2{B}-\cot^2{A}}\) ➜Note \(\because (a-b)(a+b)=a^2-b^2\)

\(\therefore \cot{(A+B)}\cot{(A-B)}=\frac{\cot^2{A}\cot^2{B}-1}{\cot^2{B}-\cot^2{A}}\)
\(\cot{(A+B)}\cot{(A-B)}=\frac{\cot^2{A}\cot^2{B}-1}{\cot^2{B}-\cot^2{A}}\)

প্রমাণঃ
আমরা জানি,
\(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
\(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
এখন, \(\sin{(A+B+C)}=\sin{\{(A+B)+C\}}\)
\(=\sin{(A+B)}\cos{C}+\cos{(A+B)}\sin{C}\)
\(=(\sin{A}\cos{B}+\cos{A}\sin{B})\cos{C}+(\cos{A}\cos{B}-\sin{A}\sin{B})\sin{C}\) ➜Note \(\because \sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
এবং \(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)

\(=\sin{A}\cos{B}\cos{C}+\cos{A}\sin{B}\cos{C}+\cos{A}\cos{B}\sin{C}-\sin{A}\sin{B}\sin{C}\)
\(=\cos{A}\cos{B}\cos{C}\left(\frac{\sin{A}}{\cos{A}}+\frac{\sin{B}}{\cos{B}}+\frac{\sin{C}}{\cos{C}}-\frac{\sin{A}\sin{B}\sin{C}}{\cos{A}\cos{B}\cos{C}}\right)\)
\(=\cos{A}\cos{B}\cos{C}(\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C})\) ➜Note \(\because \frac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}\)

\(\therefore \sin{(A+B+C)}=\cos{A}\cos{B}\cos{C}(\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C})\)
\(\sin{(A+B+C)}=\cos{A}\cos{B}\cos{C}(\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C})\)

প্রমাণঃ
আমরা জানি,
\(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
\(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
এখন, \(\cos{(A+B+C)}=\cos{\{(A+B)+C\}}\)
\(=\cos{(A+B)}\cos{C}-\sin{(A+B)}\sin{C}\)
\(=(\cos{A}\cos{B}-\sin{A}\sin{B})\cos{C}+(\sin{A}\cos{B}+\cos{A}\sin{B})\sin{C}\) ➜Note \(\because \cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
এবং \(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)

\(=\cos{A}\cos{B}\cos{C}-\sin{A}\sin{B}\cos{C}+\sin{A}\cos{B}\sin{C}+\cos{A}\sin{B}\sin{C}\)
\(=\cos{A}\cos{B}\cos{C}\left(1-\frac{\sin{A}\sin{B}}{\cos{A}\cos{B}}-\frac{\sin{B}\sin{C}}{\cos{B}\cos{C}}-\frac{\sin{C}\sin{A}}{\cos{C}\cos{A}}\right)\)
\(=\cos{A}\cos{B}\cos{C}(1-\tan{A}\tan{B}-\tan{B}\tan{C}-\tan{C}\tan{A})\) ➜Note \(\because \frac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}\)

\(\therefore \cos{(A+B+C)}=\cos{A}\cos{B}\cos{C}(1-\tan{A}\tan{B}-\tan{B}\tan{C}-\tan{C}\tan{A})\)
\(\cos{(A+B+C)}=\cos{A}\cos{B}\cos{C}(1-\tan{A}\tan{B}-\tan{B}\tan{C}-\tan{C}\tan{A})\)

প্রমাণঃ
আমরা জানি,
\(\sin{(A+B+C)}=\cos{A}\cos{B}\cos{C}(\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C})\)
\(\cos{(A+B+C)}=\cos{A}\cos{B}\cos{C}(1-\tan{A}\tan{B}-\tan{B}\tan{C}-\tan{C}\tan{A})\)
এখন, \(\tan{(A+B+C)}=\frac{\sin{(A+B+C)}}{\cos{(A+B+C)}}\)
\(=\frac{\cos{A}\cos{B}\cos{C}(\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C})}{\cos{A}\cos{B}\cos{C}(1-\tan{A}\tan{B}-\tan{B}\tan{C}-\tan{C}\tan{A})}\)
\(=\frac{\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C}}{1-\tan{A}\tan{B}-\tan{B}\tan{C}-\tan{C}\tan{A}}\)
\(\therefore \tan{(A+B+C)}=\frac{\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C}}{1-\tan{A}\tan{B}-\tan{B}\tan{C}-\tan{C}\tan{A}}\)
\(\tan{(A+B+C)}=\frac{\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C}}{1-\tan{A}\tan{B}-\tan{B}\tan{C}-\tan{C}\tan{A}}\)

প্রমাণঃ
লেখা যায়,
\(\cot{(A+B+C)}=\cot{\{(A+B)+C\}}\)
\(=\frac{\cot{(A+B)}\cot{C}-1}{\cot{C}+\cot{(A+B)}}\) ➜Note \(\because \cot{(P+Q)}=\frac{\cot{P}\cot{Q}-1}{\cot{Q}+\cot{P}}\)

\(=\frac{\frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}}\cot{C}-1}{\cot{C}+\frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}}}\) ➜Note \(\because \cot{(P+Q)}=\frac{\cot{P}\cot{Q}-1}{\cot{Q}+\cot{P}}\)

\(=\frac{\frac{\cot{A}\cot{B}\cot{C}-\cot{C}}{\cot{B}+\cot{A}}-1}{\cot{C}+\frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}}}\)
\(=\frac{\cot{A}\cot{B}\cot{C}-\cot{C}-\cot{B}-\cot{A}}{\cot{B}\cot{C}+\cot{C}\cot{A}+\cot{A}\cot{B}-1}\) ➜Note লব ও হরকে \((\cot{B}+\cot{A})\) দ্বারা গুণ করে।

\(\therefore \cot{(A+B+C)}=\frac{\cot{A}\cot{B}\cot{C}-\cot{A}-\cot{B}-\cot{C}}{\cot{B}\cot{C}+\cot{C}\cot{A}+\cot{A}\cot{B}-1}\)

\(\cot{(A+B+C)}=\frac{\cot{A}\cot{B}\cot{C}-\cot{A}-\cot{B}-\cot{C}}{\cot{B}\cot{C}+\cot{C}\cot{A}+\cot{A}\cot{B}-1}\)