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 660: 69

Answer

The expression in terms of the function $\cos x$ is $\frac{1}{\cos x}$.

Work Step by Step

By using the quotient identity of trigonometry $\tan x=\frac{\sin x}{\cos x}$. The above expression can be further solved as, $\frac{\cos x}{1+\sin x}+\tan x=\frac{\cos x}{1+\sin x}+\frac{\sin x}{\cos x}$ Now, simplify the expression on the basis of the common factor. Then we get, $\begin{align} & \frac{\cos x}{1+\sin x}+\frac{\sin x}{\cos x}=\frac{{{\cos }^{2}}x+\sin x+{{\sin }^{2}}x}{\left( 1+\sin x \right)\left( \cos x \right)} \\ & =\frac{{{\cos }^{2}}x+{{\sin }^{2}}x+\sin x}{\left( 1+\sin x \right)\left( \cos x \right)} \end{align}$ Now, apply the Pythagorean identity of trigonometry ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Then the above expression can be simplified as, $\begin{align} & \frac{{{\cos }^{2}}x+{{\sin }^{2}}x+\sin x}{\left( 1+\sin x \right)\left( \cos x \right)}=\frac{1+\sin x}{\left( 1+\sin x \right)\left( \cos x \right)} \\ & =\frac{1}{\cos x} \end{align}$ Thus, the provided expression is written in the terms of the function $\cos x$ as $\frac{1}{\cos x}$. Hence, the expression $\frac{\cos x}{1+\sin x}+\tan x$ in terms of the function $\cos x$ is $\frac{1}{\cos x}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.