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: 33

Answer

$${\text{sec}}\left( {\frac{\pi }{2} - \theta } \right) = \csc \theta $$

Work Step by Step

$$\eqalign{ & {\text{Let }}y = {\text{sec}}\left( {\frac{\pi }{2} - \theta } \right) \cr & {\text{Use the identity }}\sec \alpha = \frac{1}{{\cos \alpha }},{\text{ then}} \cr & y = \frac{1}{{\cos \left( {\frac{\pi }{2} - \theta } \right)}} \cr & {\text{Where cos}}\left( {A - B} \right) = \cos A\cos B + \sin A\sin B \cr & y = \frac{1}{{\cos \left( {\frac{\pi }{2}} \right)\cos \left( \theta \right) + \sin \left( {\frac{\pi }{2}} \right)\sin \left( \theta \right)}} \cr & {\text{Simplify}} \cr & y = \frac{1}{{\left( 0 \right)\cos \left( \theta \right) + 1 \sin \left( \theta \right)}} \cr & y = \frac{1}{{\sin \theta }} \cr & y = \csc \theta \cr & {\text{Therefore,}} \cr & {\text{sec}}\left( {\frac{\pi }{2} - \theta } \right) = \csc \theta \cr} $$
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