Precalculus (6th Edition) Blitzer

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

Chapter 7 - Section 7.5 - Systems of Inequalities - Exercise Set - Page 866: 119


See the verification below.

Work Step by Step

The left side of the identity is as given below: $\frac{1}{\sin x\times \cos x}-\frac{\cos x}{\sin x}$ Simplifying the left side gives: $\begin{align} & \frac{\sec x}{\sin x}-\frac{\cos x}{\sin x}=\frac{\sec x-\cos x}{\sin x} \\ & =\frac{\frac{1}{\cos x}-\cos x}{\sin x} \\ & =\frac{1-{{\cos }^{2}}x}{\sin x\times \cos x} \end{align}$ Simplifying further, $\begin{align} & \frac{1-{{\cos }^{2}}x}{\sin x\times \cos x}=\frac{{{\sin }^{2}}x}{\sin x\times \cos x} \\ & =\frac{\sin x}{\cos x} \\ & =\tan x \end{align}$ Since, $\text{left side = right side}$ , we can say that the given identity is true. Thus, it is verified.
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