Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 2 - Section 2.4 - Applications - 2.4 Problem Set - Page 97: 50

Answer

Please refer to the step-by-step [art below for the solution.

Work Step by Step

RECALL: $\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} \\\csc{\theta} = \dfrac{1}{\sin{\theta}}$ Thus, the left side of the equation is equivalent to: $\require{cancel} \cos{\theta} \csc{\theta}\tan{\theta}$ $\\=\cos{\theta} \cdot \dfrac{1}{\sin{\theta}} \cdot \dfrac{\sin{\theta}}{\cos{\theta}} \\=\cancel{\cos{\theta}} \cdot \dfrac{1}{\cancel{\sin{\theta}}} \cdot \dfrac{\cancel{\sin{\theta}}}{\cancel{\cos{\theta}}} \\=1.$ Therefore, $\cos{\theta} \csc{\theta}\tan{\theta}=1$
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