Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Mid-Chapter Check Point - Page 684: 20



Work Step by Step

Step 1. Given $sin\alpha=\frac{3}{5}, \frac{\pi}{2}\lt\alpha\lt\pi$, we know $\alpha$ is in quadrant II; thus $cos\alpha=-\sqrt {1-sin^2\alpha}=-\frac{4}{5}$ and $tan\alpha=-\frac{3}{4}$ Step 2. Given $cos\beta=-\frac{12}{13}, \pi\lt\beta\lt \frac{3\pi}{2}$, we know $\beta$ is in quadrant III; thus $sin\beta=-\sqrt {1-cos^2\beta}=-\frac{5}{13}$ and $tan\beta=\frac{5}{12}$ Step 3. Using the Addition Formula, we have $tan(\alpha+\beta)=\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta }=\frac{(-\frac{3}{4})+(\frac{5}{12})}{1-(-\frac{3}{4})(\frac{5}{12})}=-\frac{16}{63}$
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.