Question

If $\tan A=k\tan B$, show that,
$\sin \left(A+B\right)=\left(\frac{k+1}{k-1}\right)\sin \left(A-B\right)$

Answer 1

The question must be to show that $\sin (A+B)=\frac{k+1}{k-1}\sin (A-B)$

$\tan A=k\tan B$

$\frac{\tan A}{\tan B}=\frac{k}{1}$

$\Rightarrow \frac{\tan A+\tan B}{\tan A-\tan B}=\frac{k+1}{k-1}$

$\Rightarrow \frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}=\frac{k+1}{k-1}$

$\Rightarrow \frac{\sin A\cos B+\cos A\sin B}{\sin A\cos B-\cos A\sin B}=\frac{k+1}{k-1}$

$\Rightarrow \frac{\sin (A+B)}{\sin (A-B)}=\frac{k+1}{k-1}$

$\Rightarrow \sin (A+B)=\frac{k+1}{k-1}\sin (A-B)$