Precalculus (6th Edition) Blitzer

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

Chapter 7 - Cumulative Review Exercises - Page 880: 36


See the explanation given below.

Work Step by Step

Let us consider the left side of the expression: $\begin{align} & \sec \theta -\cos \theta =\frac{1}{\cos \theta }-\frac{\cos \theta }{1} \\ & =\frac{1-{{\cos }^{2}}\theta }{\cos \theta } \\ & =\frac{{{\sin }^{2}}\theta }{\cos \theta } \\ & =\frac{\sin \theta }{\cos \theta }\sin \theta \end{align}$ Finally, use the identities $\sec \theta =\frac{1}{\cos \theta },\ \tan \theta =\frac{\sin \theta }{\cos \theta }$. Therefore, $\sec \theta -\cos \theta =\tan \theta \sin \theta $ Thus, the left side of the expression is equal to right side.
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