Question

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

Answer 1

Given: $\tan A=x\tan B$

To prove: $\frac{\sin (A-B)}{\sin (A+B)}=\frac{x-1}{x+1}$

Now,
$\frac{\sin (A-B)}{\sin (A+B)}$

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

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

$=\frac{\frac{\tan A}{\tan B}-1}{\frac{\tan A}{\tan B}+1}$ $[\because \tan A=x\tan B]$

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

$\text{Hence proved.}$