Answer
The graphs appear to coincide.
Work Step by Step
Please note that$$\tan (\alpha - \beta ) = \frac{\tan \alpha - \tan \beta }{1+\tan \alpha \tan \beta }.$$So we have$$\tan (\pi -x)=\frac{\tan \pi - \tan x}{1+\tan \pi \tan x}=-\tan x .$$
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
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 671: 89
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
