Trigonometry (10th Edition)

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

Chapter 5 - Trigonometric Identities - Section 5.1 Fundamental Identities - 5.1 Exercises: 58

Answer

$sec~x = \pm \frac{\sqrt{1 - sin^2~x}}{1 - sin^2~x}$

Work Step by Step

We can find an expression for $cos~x$ in terms of $sin~x$: $sin^2~x+cos^2~x = 1$ $cos^2~x = 1 - sin^2~x$ $cos~x = \pm~\sqrt{1 - sin^2~x}$ We can write $sec~x$ in terms of $sin~x$: $sec~x = \frac{1}{cos~x}$ $sec~x = \frac{1}{\pm~\sqrt{1 - sin^2~x}}$ $sec~x = \frac{1}{\pm~\sqrt{1 - sin^2~x}}~\frac{\sqrt{1 - sin^2~x}}{\sqrt{1 - sin^2~x}}$ $sec~x = \pm \frac{\sqrt{1 - sin^2~x}}{1 - sin^2~x}$
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