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: 45


The left side of the equation is equivalent to $\dfrac{\cot{v}+1}{\cot{v}-1}$ therefore the given equation is an identity. Refer to the solution below.

Work Step by Step

Work on the left hand side (LHS). Divide by } the nnumerator and denomoinator of the LHS by $\tan{v}$ to obtain: \begin{align} \text{LHS } & = \dfrac{\dfrac{1}{\tan{v}}+1}{\dfrac{1}{\tan{v}}-1} \\[3mm] &= \dfrac{\cot{v}+1}{\cot{v}-1} \\[3mm] &= \text{ RHS} \end{align}
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