Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.1 - Verifying Trigonometric Identities - Exercise Set - Page 659: 24

Answer

See the explanation below.

Work Step by Step

$\frac{1-\sin \theta }{\cos \theta }=\sec \theta -\tan \theta $ Recall Trigonometric Identities, $\begin{align} & \tan \theta =\frac{\sin \theta }{\cos \theta } \\ & \sec \theta =\frac{1}{\cos \theta } \\ \end{align}$ Use the above identities and solve the left side of the given expression, $\begin{align} & \frac{1-\sin \theta }{\cos \theta }=\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta } \\ & =\sec \theta -\tan \theta \end{align}$ Therefore, $\frac{1-\sin \theta }{\cos \theta }=\sec \theta -\tan \theta $
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