Question
Prove The Theorem : For any two angles A and B,
$\sin (A+B)=\sin A \cos B+\cos A \sin B \quad \text { [verify] }$
$\sin (A+B)=\sin A \cos B+\cos A \sin B \quad \text { [verify] }$
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Answer
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Similar questions
Which of the following is true?
i) $2 \cos ^2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}$
ii) $\frac{\cot A-\tan B}{\cot B-\tan A}=\cot A \tan B$
iii) $\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}=\sin \theta+\cos \theta$
→i) $2 \cos ^2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}$
ii) $\frac{\cot A-\tan B}{\cot B-\tan A}=\cot A \tan B$
iii) $\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}=\sin \theta+\cos \theta$
Theorem : For any two angles A and B,
$\sin (A-B)=\sin A \cos B-\cos A \sin B$
→$\sin (A-B)=\sin A \cos B-\cos A \sin B$
Prove the Theorem : For any angle $\theta$,
1)$\sin 2 \theta=2 \sin \theta \cos \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta}$
2)$
\begin{aligned}
& \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta=2 \cos ^2 \theta-1 \\
& =1-2 \sin ^2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}
\end{aligned}
$
3)$\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}$
→1)$\sin 2 \theta=2 \sin \theta \cos \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta}$
2)$
\begin{aligned}
& \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta=2 \cos ^2 \theta-1 \\
& =1-2 \sin ^2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}
\end{aligned}
$
3)$\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}$
Theorem : For any two angles A and B,
$\cos (A+B)=\cos A \cos B-\sin A \sin B$
→$\cos (A+B)=\cos A \cos B-\sin A \sin B$
Prove The Theorem : For any two angles A and B, $\tan (\mathrm{A}-\mathrm{B})=\frac{\tan A-\tan B}{1+\tan A \tan B}$ (Activity)
→Draw the angle ABC of measure 40°
Find trigonometric functions of angles $150^{\circ}$, $210^{\circ}, 330^{\circ},-45^{\circ},-120^{\circ},-\frac{3 \pi}{4}$ and complete the table.
→
Find trigonometric functions of angles 180°, 270°.
Differentiate $\tan x$ and $\sec x$ w.r.t. $x$ using the formulae for differentiation of $\frac{u}{v}$ and $\frac{1}{v}$ respectively.
→Prove The Theorem : For any two angles A and B, cos $(A-B)=\cos A \cos B+\sin A \sin B$
→