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: 10

Answer

$\dfrac{1}{\sin{\theta}}$

Work Step by Step

Recall: $\cot{\theta} =\dfrac{\cos{\theta}}{\sin{\theta}} \hspace{20pt} \text{and} \hspace{20pt} \sec{\theta} = \dfrac{1}{\cos{\theta}}$ Thus, $\cot{\theta} \cdot \sec{\theta}= \dfrac{\cos{\theta}}{\sin{\theta}} \times \dfrac{1}{\cos{\theta}} = \boxed{\dfrac{1}{\sin{\theta}}}$
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