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 667: 62


$\dfrac {1}{sec\alpha -\tan \alpha }=sec\alpha +\tan \alpha $

Work Step by Step

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