Answer
$\frac{2\cos 2\theta}{\sin 2\theta}=\cot\theta-\tan\theta$
Work Step by Step
Start with the right side:
$\cot\theta-\tan\theta$
Rewrite it in terms of sine and cosine:
$=\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}$
Get a common denominator:
$=\frac{\cos\theta}{\sin\theta}*\frac{\cos\theta}{\cos\theta}-\frac{\sin\theta}{\cos\theta}*\frac{\sin\theta}{\sin\theta}$
$=\frac{\cos^2\theta}{\sin\theta\cos\theta}-\frac{\sin^2\theta}{\sin\theta\cos\theta}$
$=\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}$
Use the identity $\cos^2\theta-\sin^2\theta=\cos 2\theta$:
$=\frac{\cos 2\theta}{\sin\theta\cos\theta}$
Multiply top and bottom by $2$:
$=\frac{2\cos 2\theta}{2\sin\theta\cos\theta}$
Use the identity $2\sin\theta\cos\theta=\sin 2\theta$:
$=\frac{2\cos 2\theta}{\sin 2\theta}$
Since this equals the left side, the identity has been proven.
