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 658: 8

Answer

See the explanation below.

Work Step by Step

To verify the given identity, $\csc x-\csc x{{\sin }^{2}}x=\sin x$ Recall Trigonometric Identities, $\begin{align} & \csc x=\frac{1}{\sin x} \\ & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\ \end{align}$ Use the above identities and solve the left side of the given expression, $\begin{align} & \csc x-\csc x{{\sin }^{2}}x=\csc x\left( 1-{{\cos }^{2}}x \right) \\ & =\left( \frac{1}{\sin x} \right)\left( {{\sin }^{2}}x \right) \\ & =\sin x \end{align}$ Hence, it is proved that the given identity holds true.
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