Answer
The identity is verified.
$tan(θ+\pi)=tan~θ$
Work Step by Step
$tan(u+v)=\frac{tan~u+tan~v}{1-tan~u~tan~v}$
$tan~\pi=0$
$tan(θ+\pi)=\frac{tan~θ+tan~\pi}{1-tan~θ~tan~θ}=\frac{tan~θ+0}{1-tan~θ(0)}=tan~θ$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 7 - 7.4 - Sum and Difference Equations - 7.4 Exercises - Page 539: 61
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