Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 37


See the full explanation below.

Work Step by Step

Let us consider the left side of the given expression: $\tan \left( 2\pi -x \right)$ By using the identity of trigonometry, $\tan \,\left( \alpha -\beta \right)=\frac{\tan \,\alpha -\tan \,\beta }{1+\tan \,\alpha \tan \,\beta }$, the above expression can be further simplified as: $\begin{align} & \tan \left( 2\pi -x \right)=\frac{\tan \,2\pi -\tan \,x}{1+\tan \,2\pi \tan \,x} \\ & =\frac{0-\tan \,x}{1+0\times \tan \,x} \\ & =\frac{-\tan \,x}{1} \\ & =-\tan \,x \end{align}$ Hence, the the left side of the provided expression is equal to the right side, which is $\tan \left( 2\pi -x \right)=-\tan \,x$.
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.