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: 50

Answer

$sin~\theta=\frac{\sqrt {21}}{5}$ $tan~\theta=-\frac{\sqrt {21}}{2}$ $cot~\theta=-\frac{2\sqrt {21}}{21}$ $sec~\theta=-\frac{5}{2}$ $csc~\theta=\frac{5\sqrt {21}}{21}$

Work Step by Step

$sin^2\theta+cos^2\theta=1$ $sin^2\theta+(-\frac{2}{5})^2=1$ $sin^2\theta=1-\frac{4}{25}=\frac{21}{25}$ $sin~\theta=\frac{\sqrt {21}}{5}$ $tan~\theta=\frac{sin~\theta}{cos~\theta}=\frac{\frac{\sqrt {21}}{5}}{-\frac{2}{5}}=-\frac{\sqrt {21}}{2}$ $cot~\theta=\frac{cos~\theta}{sin~\theta}=\frac{-\frac{2}{5}}{\frac{\sqrt {21}}{5}}=-\frac{2\sqrt {21}}{21}$ $sec~\theta=\frac{1}{cos~\theta}=-\frac{5}{2}$ $csc~\theta=\frac{1}{sin~\theta}=\frac{5}{\sqrt {21}}=\frac{5\sqrt {21}}{21}$
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