Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.1 Fundamental Identities - 7.1 Exercises - Page 659: 52


$\dfrac {1}{\pm \sqrt {1-\sin ^{2}x}}=\pm \dfrac {\sqrt {1-\sin ^{2}x}}{1-\sin ^{2}x}$

Work Step by Step

$secx=\dfrac {1}{\cos x}=\dfrac {1}{\pm \sqrt {1-\sin ^{2}x}}=\pm \dfrac {\sqrt {1-\sin ^{2}x}}{1-\sin ^{2}x}$
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.