Answer
$\frac{2\tan B}{\sin 2B}=\sec^2 B$
Work Step by Step
Start with the left side:
$\frac{2\tan B}{\sin 2B}$
Rewrite in terms of $\sin B$ and $\cos B$:
$=\frac{2*\frac{\sin B}{\cos B}}{2\sin B\cos B}$
Multiply top and bottom by $\cos B$:
$=\frac{2*\frac{\sin B}{\cos B}}{2\sin B\cos B}*\frac{\cos B}{\cos B}$
$=\frac{2\sin B}{2\sin B\cos^2 B}$
Simplify:
$=\frac{1}{\cos^2 B}$
$=\sec^2 B$
