Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Cumulative Review Exercises - Page 1038: 21

Answer

See explanations.

Work Step by Step

Step 1. $LHS=\frac{\frac{1}{sin\theta}-sin\theta}{sin\theta}=\frac{1-sin^2\theta}{sin^2\theta}=\frac{cos^2\theta}{sin^2\theta}$ Step 2. $RHS=cot^2\theta=\frac{cos^2\theta}{sin^2\theta}$ Step 3. Since $LHS=RHS$, we verified the identify.
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