Prove that $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$

solution:

It is known that

$\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \cos ^{2} A-\sin ^{2} A=\cos 2 A$

$\therefore$ L.H.S. $=\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$

$=\frac{2 \cos \left(\frac{x+3 x}{2}\right) \sin \left(\frac{x-3 x}{2}\right)}{-\cos 2 x}$

$=\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}$

$=-2 \times(-\sin x)$

$=2 \sin x=$ R.H.S.