Precalculus (10th Edition)

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

Chapter 8 - Applications of Trigonometric Functions - Chapter Review - Review Exercises - Page 557: 4



Work Step by Step

We know that $\sec(\theta)=\csc(90^{\circ}-\theta)$. Hence, $\frac{\sec(55^{\circ})}{\csc(35^{\circ})}\\ =\frac{\sec(55^{\circ})}{\csc(90^{\circ}-35^{\circ})}\\ =\frac{\sec(55^{\circ})}{\sec(55^{\circ})}\\ =1$
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