Answer
$\dfrac {1}{\pm \sqrt {1-\sin ^{2}x}}=\pm \dfrac {\sqrt {1-\sin ^{2}x}}{1-\sin ^{2}x}$
Work Step by Step
$secx=\dfrac {1}{\cos x}=\dfrac {1}{\pm \sqrt {1-\sin ^{2}x}}=\pm \dfrac {\sqrt {1-\sin ^{2}x}}{1-\sin ^{2}x}$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
Published by
Pearson
ISBN 10:
013421742X
ISBN 13:
978-0-13421-742-0
Chapter 7 - Trigonometric Identities and Equations - 7.1 Fundamental Identities - 7.1 Exercises - Page 659: 52
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