Question:
If $\tan A=x \tan B$, prove that $\frac{\sin (A-B)}{\sin (A+B)}=\frac{x-1}{x+1}$.
Solution:
LHS $=\frac{\sin (A-B)}{\sin (A+B)}$
$=\frac{\sin A \cos B-\cos A \sin B}{\sin A \cos B+\cos A \sin B}$
Dividing numerator and denominator by $\cos A \cos B$ :
$\frac{\tan A-\tan B}{\tan A+\tan B}$
$=\frac{x \tan B-\tan B}{x \tan B+\tan B} \quad($ Since $\tan A=x \tan B)$
$=\frac{\tan B(x-1)}{\tan B(x+1)}$
$=\frac{x-1}{x+1}$
= RHS
Hence proved.
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