#### Answer

$\tan 2\theta=\frac{-2\tan\theta}{\sec^2\theta-2}$

#### Work Step by Step

Start with the right side:
$\frac{-2\tan\theta}{\sec^2\theta-2}$
Rewrite $\sec^2\theta$ as $\tan^2\theta+1$:
$=\frac{-2\tan\theta}{\tan^2\theta+1-2}$
$=\frac{-2\tan\theta}{\tan^2\theta-1}$
Multiply top and bottom by $-1$:
$=\frac{-2\tan\theta}{\tan^2\theta-1}*\frac{-1}{-1}$
$=\frac{2\tan\theta}{1-\tan^2\theta}$
Use the identity $\frac{2\tan\theta}{1-\tan^2\theta}=\tan 2\theta$:
$=\tan 2\theta$
Since this equals the left side, the identity has been proven.