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: 46


$cot~\theta=-\frac{\sqrt {5}}{2}$ $tan~\theta=-\frac{2\sqrt {5}}{5}$ $sec^2\theta=-\frac{3\sqrt {5}}{5}$ $cos~\theta=-\frac{\sqrt 5}{3}$ $sin~\theta=\frac{2}{3}$

Work Step by Step

$csc^2\theta=1+cot^2\theta$ $cot^2\theta=(\frac{3}{2})^2-1=\frac{5}{4}$ $cot~\theta=-\frac{\sqrt {5}}{2}~~~~$ $(sin~\theta\gt0$ and $cos~\theta\lt0)$ $tan~\theta=\frac{1}{cot~\theta}=-\frac{2}{\sqrt {5}}=-\frac{2\sqrt {5}}{5}$ $sec^2\theta=tan^2\theta+1=\frac{4}{5}+1=\frac{9}{5}$ $sec~\theta=-\frac{3}{\sqrt {5}}=-\frac{3\sqrt {5}}{5}$ $cos~\theta=\frac{1}{sec~\theta}=-\frac{\sqrt 5}{3}$ $sin~\theta=\frac{1}{csc~\theta}=\frac{2}{3}$
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