Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.4 Trigonometric Identities - 6.4 Assess Your Understanding - Page 496: 9

Answer

$\dfrac{1}{\cos{\theta}} $

Work Step by Step

Since $\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} \hspace{20pt} \text{and} \hspace{20pt} \csc{\theta} = \dfrac{1}{\sin{\theta}}$ Then $\tan{\theta} \cdot \csc{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} \cdot \dfrac{1}{\sin{\theta}} = \boxed{\dfrac{1}{\cos{\theta}} }$
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