Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 1 - Trigonometric Functions - Section 1.4 Using the Definitions of the Trigonometric Functions - 1.4 Exercises: 64

Answer

$\sec \theta = -\frac{4}{3}$

Work Step by Step

Given $\tan\theta = \frac{\sqrt 7}{3}$ and $\theta $ is in quadrant III. Using Identity, $\sec^{2}\theta = 1 + \tan^{2}\theta$ $\sec^{2}\theta = 1 + (\frac{\sqrt 7}{3})^{2}$ $ = 1 + (\frac{ 7}{9})$ $= \frac{9+7}{9} = \frac{16}{9}$ $\sec^{2}\theta = \frac{16}{9}$ $\sec \theta = +\frac{4}{3} $ or $-\frac{4}{3} $ In quadrant III $\sec \theta $ is negative. Therefore, $\sec \theta =-\frac{4}{3} $
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