Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.2 Verifying Trigonometric Identities - 7.2 Exercises - Page 666: 22

Answer

$-2sec^2\alpha $

Work Step by Step

$\dfrac {1}{\sin \alpha -1}-\dfrac {1}{\sin \alpha +1}=\dfrac {\left( \sin \alpha +1\right) -\left( \sin \alpha -1\right) }{\left( \sin \alpha -1\right) \left( \sin \alpha +1\right) }=\dfrac {2}{-\left( 1-\sin ^{2}\alpha \right) }=\dfrac {2}{-\cos ^{2}\alpha }=-2sec^2\alpha $
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.