Precalculus (6th Edition)

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: 51


$\dfrac {1}{\pm \sqrt {1-\cos ^{2}x}}=\pm \dfrac {\sqrt {1-\cos ^{2}x}}{1-\cos ^{2}x}$

Work Step by Step

$cscx=\dfrac {1}{\sin x}=\dfrac {1}{\pm \sqrt {1-\cos ^{2}x}}=\pm \dfrac {\sqrt {1-\cos ^{2}x}}{1-\cos ^{2}x}$
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