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: 47


See the explanation below.

Work Step by Step

$\frac{\tan t}{\sec t-1}$ We multiply $\frac{\sec t+1}{\sec t+1}$ with the numerator and denominator of the provided expression: $\begin{align} & \frac{\tan t}{\sec t-1}=\frac{\tan t}{\sec t-1}.\frac{\sec t+1}{\sec t+1} \\ & =\frac{\tan t\left( \sec t+1 \right)}{{{\sec }^{2}}t-1} \end{align}$ Applying the Pythagorean identity of trigonometry ${{\tan }^{2}}t={{\sec }^{2}}t-1$ , which comes out by solving $1+{{\tan }^{2}}t={{\sec }^{2}}t$ , then the above expression can be further simplified as: $\begin{align} & \frac{\tan t\left( \sec t+1 \right)}{{{\sec }^{2}}t-1}=\frac{\tan t\left( \sec t+1 \right)}{{{\tan }^{2}}t} \\ & =\frac{\sec t+1}{\tan t} \end{align}$ Thus, the right side of the expression is equal to the left side, which is $\frac{\sec t+1}{\tan t}=\frac{\tan t}{\sec t-1}$.
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