Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.4 Trigonometric Functions and Their Inverse - 1.4 Exercises - Page 48: 31

Answer

$\textbf{See Step By Step for Proof.}$

Work Step by Step

$tan^2(\theta) + 1 = sec^2(\theta)$ $\frac{sin^2(\theta)}{cos^2(\theta)}+1 = sec^2(\theta)$ $\frac{sin^2(\theta)+cos^2(\theta)}{cos^2(\theta)} = sec^2(\theta)$ $\dfrac{1}{cos^2(\theta)} = sec^2(\theta)$ $sec^2(\theta) = sec^2(\theta)$
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