Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.5 Sum and Difference Formulas - 6.5 Assess Your Understanding - Page 509: 53

Answer

See proof below.

Work Step by Step

Apply the difference formula: $\tan(A -B)=\dfrac{\tan A - \tan B}{1 +\tan A \ \tan B }$ In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows: $\text{LHS } =\tan(\pi + \theta) \\=\dfrac{\tan (\pi) -\tan (\theta)}{1+ \tan (\pi) \ \tan (\theta)} \\=\dfrac{0-\tan (\theta)}{1+ (0) \ \tan (\theta)} \\=-\tan (\theta)\\=\text{ RHS}$
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.