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: 27


See the explanation below.

Work Step by Step

$\tan t+\frac{\cos t}{1+\sin t}=\sec t$ Recall Trigonometric Identities, $\begin{align} & \tan t=\frac{\sin t}{\cos t} \\ & {{\sin }^{2}}t+{{\cos }^{2}}t=1 \\ \end{align}$ Use the above identities and solve the left side of the given expression, $\tan t+\frac{\cos t}{1+\sin t}=\frac{\sin t}{\cos t}+\frac{\cos t}{1+\sin t}$ Multiply numerator and denominator of $\tan t$ by $1+\sin t$ and $\frac{\cos t}{1+\sin t}$ by $\cos t$. $\begin{align} & \tan t+\frac{\cos t}{1+\sin t}=\frac{\sin t}{\cos t}\left( \frac{1+\sin t}{1+\sin t} \right)+\frac{\cos t}{1+\sin t}\left( \frac{\cos t}{\cos t} \right) \\ & =\frac{\sin t+{{\sin }^{2}}t}{\cos t+\cos t\sin t}+\frac{{{\cos }^{2}}t}{\cos t+\sin t\cos t} \\ & =\frac{\sin t+1}{\cos t\left( 1+\sin t \right)} \\ & =\frac{1}{\cos t} \end{align}$ Recall Reciprocity Identity, $\sec t=\frac{1}{\cos t}$ Therefore, $\begin{align} & \tan t+\frac{\cos t}{1+\sin t}=\frac{1}{\cos t} \\ & =\sec t \end{align}$
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