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

(N/A) We know the trigonometric identities: $\sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$ and $\cos^2 A - \sin^2 A = \cos 2A$.
$L.H.S. = \frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x}$
$= \frac{2 \cos \left( \frac{x+3x}{2} \right) \sin \left( \frac{x-3x}{2} \right)}{-(\cos^2 x - \sin^2 x)}$
$= \frac{2 \cos(2x) \sin(-x)}{-\cos 2x}$
Since $\sin(-x) = -\sin x$,we have:
$= \frac{2 \cos 2x (-\sin x)}{-\cos 2x}$
$= 2 \sin x = R.H.S.$