Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 31: 53


$$\tan (\pi-\theta) =-\tan (\theta) $$

Work Step by Step

We use the know identities to obtain: \begin{align*} \tan (\pi-\theta)&=\frac{\sin (\pi-\theta)}{\cos (\pi-\theta)} \\ &=\frac{\sin (\pi) \cos (\theta)-\cos (\pi) \sin (\theta)}{\cos (\pi) \cos (\theta)+\sin (\pi) \sin (\theta)}\\ &=\frac{0 \times \cos (\theta)-(-1) \times \sin (\theta)}{(-1) \times \cos (\theta)+0 \times \sin (\theta)}\\ &=\frac{0+\sin (\theta)}{-\cos (\theta)+0}=\frac{\sin (\theta)}{-\cos (\theta)}\\ &=-\tan (\theta) \end{align*}
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