Prove that $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$
Prove that $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$
It is known that
$\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \cos ^{2} A-\sin ^{2} A=\cos 2 A$
$\therefore$ $L.H.S.$ $=\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$
$=\frac{2 \cos \left(\frac{x+3 x}{2}\right) \sin \left(\frac{x-3 x}{2}\right)}{-\cos 2 x}$
$=\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}$
$=-2 \times(-\sin x)$
$=2 \sin x= R . H.S.$
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