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: 46

Answer

Identity proved.

Work Step by Step

In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows: Multiply the numerator and denominator of the LHS by $\sin \ v$ to obtain: $\text{LHS } =\dfrac{\sin v(\csc v -1) }{\sin v (\csc v +1)} \\= \dfrac{\csc v \sin v -\sin v }{\csc v \sin v +\sin v}$ Since, $\csc v \sin v =\dfrac{1}{\sin v } \cdot {\sin v}=1$ Therefore, $\dfrac{\csc v \sin v -\sin v }{\csc v \sin v +\sin v} = \dfrac{1-\sin \ v}{ 1+\sin v } \\= \text{ RHS}$
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