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: 49

Answer

Refer to the proof below.

Work Step by Step

In order to prove the given identity, we simplify the left hand side $\text{LHS}$. We know that $\sin \theta =\dfrac{1}{\csc \theta}$ $\text{LHS } =\dfrac{1+\dfrac{1}{\csc \theta }}{1-\dfrac{1}{\csc \theta} } \\=\dfrac{\dfrac{\csc \theta +1}{\csc \theta}}{\dfrac{\csc \theta -1}{\csc \theta}} \\= \dfrac{\csc \theta +1}{\csc \theta -1}\\= \\= \text{ RHS}$ Thus the left-hand side equals the right-hand side and we have proven the identity.
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