# Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises: 61

$\tan\theta\cos^2\theta=\frac{2\tan\theta\cos^2\theta-\tan \theta}{1-\tan^2\theta}$

#### Work Step by Step

Start with the right side: $\frac{2\tan\theta\cos^2\theta-\tan \theta}{1-\tan^2\theta}$ Factor out $\tan \theta$ from the numerator: $=\frac{\tan\theta(2\cos^2\theta-1)}{1-\tan^2\theta}$ Simplify the denominator: $=\frac{\tan\theta(2\cos^2\theta-1)}{\frac{\cos^2\theta}{\cos^2\theta}-\frac{\sin^2\theta}{\cos^2\theta}}$ $=\frac{\tan\theta(2\cos^2\theta-1)}{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}}$ Multiply top and bottom by $\cos^2 \theta$: $=\frac{\tan\theta(2\cos^2\theta-1)}{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}}*\frac{\cos^2\theta}{\cos^2\theta}$ $=\frac{\tan\theta(2\cos^2\theta-1)\cos^2\theta}{\cos^2\theta-\sin^2\theta}$ Use the double-angle identities for cosine: $=\frac{\tan\theta\cos 2\theta\cos^2\theta}{\cos 2\theta}$ Simplify: $=\frac{\tan\theta\cos^2\theta}{1}$ $=\tan\theta\cos^2\theta$ Since this equals the left side, the identity has been proven.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.