Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 485: 58



Work Step by Step

Let $u=arccos(-\frac{3}{4})$. Then: $cos~u=-\frac{3}{4}$ The range $arccos~x$ is $0\leq x\leq\pi$. So, since $arccos(-\frac{3}{4})\gt\frac{\pi}{2}$, then $\frac{\pi}{2}\lt u\lt\pi$ (Second Quadrant) $sec[arccos(-\frac{3}{4})]=sec~u=\frac{1}{cos~u}=-\frac{4}{3}$
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