Question

Prove that $\frac{\sin 5x+\sin x-2\sin 3x}{\cos 5x-\cos x}=\tan x$

Answer 1

$LHS$

$\frac{\sin 5x+sinx-2sin3x}{\cos 5x-\mathrm{cosx}}$

$=\frac{2\sin \left(\frac{5x+x}{2}\right)\cos \left(\frac{5x-x}{2}\right)-2\sin 3x}{2\sin \left(\frac{5x+x}{2}\right)\sin \left(\frac{x-5x}{2}\right)}$

$=\frac{2\sin 3x\cos 2x-2\sin 3x}{2\sin 3x\sin (-2x)}\therefore \sin (-\theta )=-\sin \theta$

$=\frac{2\sin 3x(\cos 2x-1)}{-2\sin 3x\sin 2x}$

$=\frac{1-\cos 2x}{\sin 2x}$

$=\frac{1-(1-2{\sin }^{2}x)}{2\sin x\cos x}$

$=\frac{1-1+2{\sin }^{2}x}{2\sin x\cos x}$

$=\frac{2{\sin }^{2}x}{2\sin x\cos x}$

$=\frac{\sin x}{\cos x}$

$=\tan xRHS$

Hence, proved.