Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 16



Work Step by Step

Note that $\tan^{-1}{\left(\sqrt3\right)}=\dfrac{\pi}{3}$, because $\tan{\left(\dfrac{\pi}{3}\right)}=\sqrt3$ and $\dfrac{\pi}{3}$ is in the range of $\tan^{-1}{x}$, which is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$ Thus $\sec{\left(\tan^{-1}{\left(\sqrt3\right)}\right)}=\sec{\left(\dfrac{\pi}{3}\right)}=2.$
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