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: 51

Answer

Identity proved.

Work Step by Step

In order to prove the given identity, we simplify the left hand side $\text{LHS}$. $\text{LHS } =\dfrac{(1-\sin v)(1-\sin v) +\cos v (\cos v)}{\cos v (1-\sin v)} \\=\dfrac{1-2 \sin v+\sin^2 v+\cos^2 v}{\cos v (1-\sin v)} $ Use the identity $\sin^2 v+\cos^2 v=1$ $=\dfrac{2-2 \sin v}{\cos v (1-\sin v)} \\= \dfrac{2 (1-\sin v)}{\cos v (1-\sin v) } \\=\dfrac{2}{\cos v}\\ =2 \sec v~~ [ \because \sec v =\dfrac{1}{\cos v} ] \\ = \text{ RHS}$ Thus the left-hand side equals the right-hand side and we have proven the identity.
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