Prove that $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$
Prove that $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$
We have
${\text{L}}{\text{.H}}{\text{.S}}{\text{.}}\frac{{\sin 5x - 2\sin 3x + \sin x}}{{\cos 5x - \cos x}}$
$ = \frac{{\sin 5x + \sin x - 2\sin 3x}}{{\cos 5x - \cos x}}$
$ = \frac{{2\sin 3x\cos 2x - 2\sin 3x}}{{ - 2\sin 3x\sin 2x}}$
$ = - \frac{{\sin 3x(\cos 2x - 1)}}{{\sin 3x\sin 2x}}$
$ = \frac{{1 - \cos 2x}}{{\sin 2x}}$
$ = \frac{{2{{\sin }^2}x}}{{2\sin x\cos x}}$
$ = \tan \,x = R.H.S$
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