Precalculus (6th Edition) Blitzer

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

Chapter 5 - Review Exercises - Page 707: 1


The left side is equal to the right side.

Work Step by Step

We have the expression on the left side: $\sec x-\cos x$, which can be simplified by using the reciprocal identity $\sec x=\frac{1}{\cos x}$. $\sec x-\cos x=\frac{1}{\cos x}-\cos x$ And the expression can be further simplified by multiplying and dividing the expression by $\cos x$. $\begin{align} & \sec x-\cos x=\frac{1}{\cos x}-\cos x \\ & =\frac{1}{\cos x}-\frac{\cos x}{1}.\frac{\cos x}{\cos x} \\ & =\frac{1-{{\cos }^{2}}x}{\cos x} \end{align}$ We know the Pythagorean theorem ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$; therefore, ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $. Thus, we apply the theorem. And as per the quotient identity $\tan x=\frac{\sin x}{\cos x}$ $\begin{align} & \frac{1-{{\cos }^{2}}x}{\cos x}=\frac{{{\sin }^{2}}x}{\cos x} \\ & =\frac{\sin x}{\cos x}.\sin x \\ & =\tan x.\sin x \end{align}$ Therefore, the left side is equal to the right side.
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.