## Precalculus (6th Edition)

Published by Pearson

# Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises: 54

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

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