Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.1 Trigonometric Functions of Acute Angles - 2.1 Exercises - Page 52: 25

Answer

$\sec$ 39$^{\circ}$ = $\csc$ 51$^{\circ}$

Work Step by Step

According to the Cofunction Identities $\sec$ $A$ = $\csc$ (90$^{\circ}$ - $A$) Therefore: $\sec$ 39$^{\circ}$ = $\csc$ (90$^{\circ}$ - 39$^{\circ}$) = $\csc$ 51$^{\circ}$ Therefore: $\sec$ 39$^{\circ}$ = $\csc$ 51$^{\circ}$
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