Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - Review Exercises - Page 53: 46

Answer

$$\frac{{\sqrt {4 - {x^2}} }}{2}$$

Work Step by Step

$$\eqalign{ & {\text{From the triangle shown bellow we have that}} \cr & \tan \theta = \frac{{{\text{Opposite side}}}}{{{\text{Adjacent side}}}} \cr & \tan \theta = \frac{{\sqrt {4 - {x^2}} }}{2} \cr & and \cr & {\text{sec}}\theta = \frac{x}{2} \cr & \theta = {\sec ^{ - 1}}\left( {\frac{x}{2}} \right) \cr & \cr & {\text{Then,}} \cr & \tan \theta = \tan \left( {{{\sec }^{ - 1}}\left( {\frac{x}{2}} \right)} \right) \cr & \tan \theta = \frac{{\sqrt {4 - {x^2}} }}{2} \cr} $$
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