Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.1 Use Trigonometry with Right Triangles - 13.1 Exercises - Skill Practice - Page 856: 10


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Work Step by Step

$\sec \theta=\dfrac{1}{\dfrac{5}{8}}=\dfrac{8}{5}$ From the Pythagorean identity, we obtain: $\sin^2 \theta +\cos^2 \theta=1\\\cos^2\theta=1-\sin^2 \theta\\\cos \theta = \sqrt 1-\sin^2\theta=\sqrt 1 -(\dfrac{5}{8})^2=\sqrt \frac{39}{64}=\dfrac{\sqrt 39}{8}$ $\csc \theta=\dfrac{1}{\dfrac{\sqrt 39}{8}}=\dfrac{8}{\sqrt 39}=\dfrac{8\sqrt 39}{39}$ $\tan \theta=\dfrac{\dfrac{\sqrt 39}{8}}{\dfrac{5}{8}}=\dfrac{\sqrt 39}{5}$ $\cot \theta=\dfrac{1}{\dfrac{\sqrt 39}{5}}=\dfrac{5}{\sqrt 39}=\dfrac{5\sqrt 39}{39}$
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