Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - Chapter Review - Review Exercises - Page 506: 28

Answer

See below.

Work Step by Step

$LHS=\frac{sin\theta}{cos\theta}\cdot\frac{cos\theta}{sin\theta}-sin^2\theta=1-sin^2\theta=cos^2\theta=RHS$
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