સાબિત કરો કે : $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$

It is known that

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

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

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

$=\frac{-2 \sin \left(\frac{9 x+5 x}{2}\right) \cdot \sin \left(\frac{9 x-5 x}{2}\right)}{2 \cos \left(\frac{17 x+3 x}{2}\right) \cdot \sin \left(\frac{17 x-3 x}{2}\right)}$

$=\frac{-2 \sin 7 x \cdot \sin 2 x}{2 \cos 10 x \cdot \sin 7 x}$

$=-\frac{\sin 2 x}{\cos 10 x}$

$=R. H.S.$