Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - Summary Exercises Verifying Trigonometric Identities - Exercises: 8

Answer

$\dfrac {2}{1+\cos x}-\tan ^{2}\dfrac {x}{2}=1$

Work Step by Step

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