Answer
$\frac{\sin 3x+\sin 7x}{\cos 3x-\cos 7x}=\cot 2x$
Work Step by Step
Start with the left side:
$\frac{\sin 3x+\sin 7x}{\cos 3x-\cos 7x}$
Use the identities $\sin x+\sin y=2\sin\frac{x+y}{2}\cos \frac{x-y}{2}$ and $\cos x-\cos y=-2\sin\frac{x+y}{2}\sin \frac{x-y}{2}$ to expand:
$=\frac{2\sin\frac{3x+7x}{2}\cos \frac{3x-7x}{2}}{-2\sin\frac{3x+7x}{2}\sin \frac{3x-7x}{2}}$
$=\frac{2\sin 5x\cos (-2x)}{-2\sin 5x\sin (-2x)}$
$=\frac{\cos (-2x)}{-\sin (-2x)}$
Use the facts that cosine is an odd function and sine is an even function to simplify:
$=\frac{\cos 2x}{-(-\sin 2x)}$
$=\frac{\cos 2x}{\sin 2x}$
$=\cot 2x$
Since this equals the right side, the identity has been proven.
