Precalculus (6th Edition) Blitzer

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

Chapter 6 - Cumulative Review Exercises - Page 801: 7


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Work Step by Step

We know that, $\sin \theta =\frac{1}{\csc \theta }$. Therefore, we can rewrite the equation: $\sin \theta \csc \theta -{{\cos }^{2}}\theta $, as: $\begin{align} & \sin \theta \csc \theta -{{\cos }^{2}}\theta =\sin \theta \left( \frac{1}{\sin \theta } \right)-{{\cos }^{2}}\theta \\ & =1-{{\cos }^{2}}\theta \\ & ={{\sin }^{2}}\theta \end{align}$ Therefore, from (1) we get; $\sin \theta \csc \theta -{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $
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