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: 47

Answer

$sec~\theta=-\frac{\sqrt {58}}{3}$ $cot~\theta=\frac{3}{7}$ $csc~\theta=-\frac{\sqrt {58}}{7}$ $cos~\theta=-\frac{3\sqrt {58}}{58}$ $sin~\theta=-\frac{7\sqrt {58}}{58}$

Work Step by Step

$sec^2\theta=1+tan^2\theta=1+(\frac{7}{3})^2=\frac{58}{9}$ $sec~\theta=-\frac{\sqrt {58}}{3}$ $cot~\theta=\frac{1}{tan~\theta}=\frac{1}{\frac{7}{3}}=\frac{3}{7}$ $csc^2\theta=cot^2\theta+1=(\frac{3}{7})^2+1=\frac{58}{49}$ $csc~\theta=-\frac{\sqrt {58}}{7}$ $cos~\theta=\frac{1}{sec~\theta}=-\frac{3}{\sqrt {58}}=-\frac{3\sqrt {58}}{58}$ $sin~\theta=\frac{1}{csc~\theta}=-\frac{7}{\sqrt {58}}=-\frac{7\sqrt {58}}{58}$
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