#### Answer

$\tan4\theta=\frac{2\tan2\theta}{2-\sec^2 2\theta}$

#### Work Step by Step

Start with the right side:
$\frac{2\tan2\theta}{2-\sec^2 2\theta}$
Write $\sec^2 2\theta$ as $1+\tan^2 2\theta$:
$=\frac{2\tan2\theta}{2-(1+\tan^2 2\theta)}$
$=\frac{2\tan2\theta}{2-1-\tan^2 2\theta}$
$=\frac{2\tan2\theta}{1-\tan^2 2\theta}$
Use the identity $\frac{2\tan A}{1-\tan^2 A}=\tan 2A$ where $A=2\theta$:
$=\tan (2*2\theta)$
$=\tan 4\theta$
Since this equals the left side, the identity has been proven.