Answer
See below.
Work Step by Step
With $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$, we have
$\sqrt {16+16tan^2\theta}=\sqrt {16(1+tan^2\theta)}=\sqrt {16(sec^2\theta)}=4sec\theta$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 6 - Analytic Trigonometry - Section 6.4 Trigonometric Identities - 6.4 Assess Your Understanding - Page 498: 105
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