Algebra and Trigonometry 10th Edition

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

Chapter 7 - Review Exercises - Page 552: 20


The identity is verified. $sec^2x~cot~x-cot~x=tan~x$

Work Step by Step

$1+tan^2x=sec^2x~~$ (Pythagorean identity): $tan^2x=sec^2x-1$ $cot~x=\frac{1}{tan~x}$ $sec^2x~cot~x-cot~x=(sec^2x-1)cot~x=tan^2x~\frac{1}{tan~x}=tan~x$
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.