Question

Prove $\frac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\frac{\sin 2x}{\cos 10x}$

Answer 1

$\cos A-\cos B=2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{B-A}{2}\right)$

$\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

$\Rightarrow$L.H.S$=\frac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}$

$\Rightarrow \frac{2\sin \left(7x\right)\sin (-2x)}{2\cos \left(10x\right)\sin \left(7x\right)}$

$\Rightarrow \frac{-2\sin \left(7x\right)\sin \left(2x\right)}{2\cos \left(10x\right)\sin 7x}$

$\Rightarrow -\frac{\sin 2x}{\cos 10x}=$R.H.S

Hence proved