Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.2 - Verifying Trigonometric Identities - 7.2 Exercises - Page 520: 24


The identity is verified. $cos~x-\frac{cos~x}{1-tan~x}=\frac{sin~x~cos~x}{sin~x-cos~x}$

Work Step by Step

Remember: $tan~x=\frac{sin~x}{cos~x}$ So: $cos~x~tan~x=cos~x~\frac{sin~x}{cos~x}=sin~x$ $cos~x-\frac{cos~x}{1-tan~x}=cos~x~\frac{1-tan~x}{1-tan~x}-\frac{cos~x}{1-tan~x}=\frac{cos~x-sin~x-cos~x}{1-tan~x}=\frac{sin~x}{tan~x-1}~\frac{cos~x}{cos~x}=\frac{sin~x~cos~x}{sin~x-cos~x}$
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.