#### Answer

$1+\tan^2\alpha=2\tan \alpha\csc 2\alpha$

#### Work Step by Step

Start with the right side:
$2\tan \alpha\csc 2\alpha$
Rewrite in terms of sine and cosine:
$=\frac{2\tan \alpha}{\sin 2\alpha}$
$=\frac{2*\frac{\sin\alpha}{\cos \alpha}}{2\sin \alpha\cos\alpha}$
Multiply top and bottom by $\cos \alpha$:
$=\frac{2*\frac{\sin\alpha}{\cos \alpha}}{2\sin \alpha\cos\alpha}*\frac{\cos\alpha}{\cos\alpha}$
$=\frac{2\sin\alpha}{2\sin \alpha\cos^2\alpha}$
Simplify:
$=\frac{1}{\cos^2 \alpha}$
$=\sec^2 \alpha$
$=1+\tan^2\alpha$
Since this equals the left side, the identity has been proven.