Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Review Exercises - Page 249: 61


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

Work Step by Step

$\frac{2~tan~\theta~cos^2~\theta-tan~\theta}{1-tan^2~\theta}$ $= (tan~\theta)~(\frac{2~cos^2~\theta-1}{1-tan^2~\theta})$ $= (tan~\theta)~\frac{2~cos^2~\theta-(sin^2~\theta+cos^2~\theta)}{\frac{cos^2~\theta-sin^2~\theta}{cos^2~\theta}}$ $= (tan~\theta)~\frac{(cos^2~\theta-sin^2~\theta)(cos^2~\theta)}{cos^2~\theta-sin^2~\theta}$ $= tan~\theta~cos^2~\theta$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.