Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 16

Answer

$2$

Work Step by Step

The range of $\tan^{-1}{x}$ is $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$. Here, $\tan^{-1}(\sqrt 3)=\dfrac{\pi}{3}$, because $\tan{(\dfrac{\pi}{3})}= \sqrt 3$ and $\dfrac{\pi}{3}$ lies in the range of $\tan^{-1}{x}$ . Therefore, we have: $\sec [\tan^{-1}(\sqrt 3)]=\sec ({\dfrac{\pi}{3}})=2$
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