Prove that

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

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

$=\frac{\cos (x+y) \sin (x-y)}{\sin (x+y) \sin (x-y)}$

$=\frac{\cos (x+y)}{\sin (x+y)}$

$=\cot (x+y)$

Using the formula,

$\cos A-\cos B=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$

$\sin A-\sin B=2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$