Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - Review Questions - Page 501: 49

Answer

$sec~\theta=-\frac{41}{9}$ $cot~\theta=-\frac{9}{40}$ $csc~\theta=\frac{41}{40}$ $cos~\theta=-\frac{9}{41}$ $sin~\theta=\frac{40}{41}$

Work Step by Step

$sec^2\theta=1+tan^2\theta=1+(-\frac{40}{9})^2=\frac{1681}{81}$ $sec~\theta=-\frac{41}{9}~~~~$ $(sin~\theta\gt0$ and $cos~\theta\lt0)$ $cot~\theta=\frac{1}{tan~\theta}=\frac{1}{-\frac{40}{9}}=-\frac{9}{40}$ $csc^2\theta=cot^2\theta+1=(-\frac{9}{40})^2+1=\frac{1681}{1600}$ $csc~\theta=\frac{41}{40}$ $cos~\theta=\frac{1}{sec~\theta}=-\frac{9}{41}$ $sin~\theta=\frac{1}{csc~\theta}=\frac{40}{41}$
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