Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.3 - Right Triangle Trigonometry - Exercise Set - Page 561: 47



Work Step by Step

Solve. $\csc 37^{\circ} \sec 53^{\circ}-\tan 53^{\circ} \cot 37^{\circ}$ We have $= \sec (90^{\circ}-37^{\circ}) \sec 53^{\circ}-\tan 53^{\circ} \tan 53^{\circ}$ or, $=\sec^2 53^{\circ}-\tan^2 53^{\circ}$ Since, $1+\tan^2 x=\sec^2 x $ So, the answer is : $=1$
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