Question

Prove that: $\frac{\sin x-\sin 3x+\sin 5x-\sin 7x}{\cos x-\cos 3x-\cos 5x+\cos 7x}=\cot 2x$

Answer 1

We have,

L.H.S.

$\frac{\sin x-\sin 3x+sin5x-sin7x}{\cos x-\cos 3x-\cos 5x+\cos 7x}$

$=\frac{(\sin x+sin5x)-(\sin 3x+sin7x)}{(\cos x-\cos 5x)-(\cos 3x-\cos 7x)}$

$=\frac{2sin\left(\frac{x+5x}{2}\right)\cos \left(\frac{x-5x}{2}\right)-2\sin \left(\frac{3x+7x}{2}\right)cos\left(\frac{3x-7x}{2}\right)}{2\sin \left(\frac{x+5x}{2}\right)\sin \left(\frac{5x-x}{2}\right)-2\sin \left(\frac{3x+7x}{2}\right)\sin \left(\frac{7x-3x}{2}\right)}$

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

$=\frac{\cos 2xsin3x-\sin 5xcos2x}{\sin 3x\sin 2x-\sin 5x\sin 2x}$

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

$=\cot 2x$

L.H.S.=R.H.S.

Hence proved.