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 497: 19

Answer

The left hand side is equivalent to $\cot{\theta}$ so the given equation is an identity. Refer to the solution below.

Work Step by Step

Recall: (1) $\csc{\theta}= \dfrac{1}{\sin{\theta}} $ (2) $\cot{\theta} = \dfrac{\cos{\theta}}{\sin{\theta}}$ Thus, working on the left hand side (LHS) of the given equation using the definitions above gives: \begin{align*} \csc{\theta} \cdot \cos{\theta} &= \dfrac{1}{\sin{\theta}} \cdot \cos{\theta}\\\\ &= \dfrac{\cos{\theta}}{\sin{\theta}}\\\\ &=\cot{\theta}\end{align*} $\therefore \text{ LHS }= \text{ RHS}$
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