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: 48

Answer

See proof below.

Work Step by Step

In order to prove the given identity, we simplify the left hand side $\text{LHS}$. Multiply by the numerator and denominator of the LHS by $\csc \theta +1$ to obtain: $\text{LHS } =\dfrac{\csc \theta -1}{\cot \theta } \cdot \dfrac{\csc \theta +1}{\csc \theta +1} \\=\dfrac{\csc^2 \theta -1}{\cot \theta (\csc \theta +1)} \\= \dfrac{\cot^2 \theta}{\cot \theta (\csc \theta +1)}\\= \dfrac{\cot \theta}{(\csc \theta +1)} \\= \text{ RHS}$
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