Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.3 Sum and Difference Identities - 7.3 Exercises - Page 679: 53



Work Step by Step

Use the subtraction formula for tangent, $\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$. $\tan\left(-\frac{5\pi}{12}\right)$ $=\tan\left(\frac{\pi}{4}-\frac{2\pi}{3}\right)$ $=\frac{\tan\frac{\pi}{4}-\tan\frac{2\pi}{3}}{1+\tan\frac{\pi}{4}\tan\frac{2\pi}{3}}$ $=\frac{1-(-\sqrt{3})}{1+1*(-\sqrt{3})}$ $=\frac{1+\sqrt{3}}{1-\sqrt{3}}$ $=\frac{1+\sqrt{3}}{1-\sqrt{3}}*\frac{1+\sqrt{3}}{1+\sqrt{3}}$ $=\frac{1+2\sqrt{3}+3}{1-3}$ $=\frac{4+2\sqrt{3}}{-2}$ $=-2-\sqrt{3}$
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.