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

solution:

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.