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: 9


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Work Step by Step

$\csc \theta=\dfrac{1}{\dfrac{5}{6}}=\dfrac{6}{5}$ From the Pythagorean identity, we obtain: $\sin^2 \theta +\cos^2 \theta=1\\\cos^2\theta=1-\sin^2 \theta\\\cos \theta = \sqrt 1-\sin^2\theta=\sqrt 1 -(\dfrac{5}{6})^2=\sqrt \frac{11}{36}=\dfrac{\sqrt 11}{6}$ $\sec \theta=\dfrac{1}{\frac{\sqrt 11}{6}}=\dfrac{6}{\sqrt 11}=\dfrac{6\sqrt 11}{11}$ $\tan \theta=\dfrac{\frac{5}{6}}{\frac{\sqrt 11}{6}}=\dfrac{5}{\sqrt 11}=\dfrac{5\sqrt 11}{11}$ $\cot \theta=\dfrac{1}{\dfrac{11}{5\sqrt 11}}=\dfrac{11}{5\sqrt 11}=\dfrac{\sqrt 11}{5}$
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