## Precalculus (6th Edition)

$\frac{2\tan B}{\sin 2B}=\sec^2 B$
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: $=\frac{1}{\cos^2 B}$ $=\sec^2 B$