Prove that: x - 3x ^2 x - ^2 x = 2 x - Vidyadip

Question

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

✓

Answer

We have,
$ \text { 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)}{\left(\cos ^2 x-\sin ^2 x\right)}$
$\quad=\frac{2 \cos \left(\frac{3 x+x}{2}\right) \sin \left(\frac{3 x-x}{2}\right)}{\cos 2 x}$
$\quad=\frac{2 \cos 2 x \sin x}{\cos 2 x}=2 \sin x $

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