Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 10

Answer

$0$

Work Step by Step

Here, $ x= -1$; $ y=0$ and $ r=\sqrt {(1)^2+(0)^2}=1$ The trigonometric ratios are as follows: $\tan \theta =\dfrac{y}{x}$ Then, we have $\tan (\pi) =\dfrac{0}{-1}=0$
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