Question

If $\tan A=x\tan B$, prove that $\frac{\sin (A-B)}{\sin (A+B)}=\frac{x-1}{x+1}$

Answer 1

$\frac{\sin (A-b)}{\sin (A+B)}=\frac{\sin A\cos B-\cos A\sin B}{\sin A\cos B+\sin B\cos A}\times \frac{\cos A}{\cos A}$

$=\frac{\tan A\cos B-\sin B}{\tan A\cos B+\sin B}\times \frac{\cos B}{\cos B}=\frac{\tan A+\tan B}{\tan A+\tan B}$

$=\frac{x\tan B-\tan B}{x\tan B+\tan B}=\frac{(x-1)}{(x+1)}\frac{\tan B}{\tan B}=\frac{x-1}{x+1}$.