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Prove that $\frac{\sin x-\sin 3 x}{\sin ^2 x-\cos ^2 x}=2 \sin x$
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Verified Answer
L.H.S. $=\frac{\sin x-\sin 3 x}{\sin ^2 x-\cos ^2 x}=\frac{-(\sin 3 x-\sin x)}{-\left(\cos ^2 x-\sin ^2 x\right)}$
$=\frac{\sin 3 x-\sin x}{\cos ^2 x-\sin ^2 x}=\frac{2 \cos \frac{3 x+x}{2} \sin \frac{3 x-x}{2}}{\cos 2 x}$
$=\frac{2 \cos 2 x \times \sin x}{\cos 2 x}\left[\because \cos 2 x=\cos ^2 x-\sin ^2 x\right]$
$=2 \sin x$
$=\frac{\sin 3 x-\sin x}{\cos ^2 x-\sin ^2 x}=\frac{2 \cos \frac{3 x+x}{2} \sin \frac{3 x-x}{2}}{\cos 2 x}$
$=\frac{2 \cos 2 x \times \sin x}{\cos 2 x}\left[\because \cos 2 x=\cos ^2 x-\sin ^2 x\right]$
$=2 \sin x$
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