Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 222: 73

Answer

Expression of $\tan\theta$ in terms of $\tan\alpha$ and $\tan\beta$. $$\tan\theta=\frac{\tan\beta-\tan\alpha}{1+\tan\beta\tan\alpha}$$

Work Step by Step

As in the previous exercise, we deduce that $$\theta=\beta-\alpha$$ which means $$\tan\theta=\tan(\beta-\alpha)$$ Apply the identity of tangent difference: $$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$ Therefore, $$\tan\theta=\frac{\tan\beta-\tan\alpha}{1+\tan\beta\tan\alpha}$$ This is the expression of $\tan\theta$ in terms of $\tan\alpha$ and $\tan\beta$.
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.