Answer
$1$
Work Step by Step
We know that $\sec(\theta)=\csc(90^{\circ}-\theta)$.
Hence,
$\frac{\sec(55^{\circ})}{\csc(35^{\circ})}\\
=\frac{\sec(55^{\circ})}{\csc(90^{\circ}-35^{\circ})}\\
=\frac{\sec(55^{\circ})}{\sec(55^{\circ})}\\
=1$
Precalculus (10th Edition)
by Sullivan, Michael
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
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