University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 181: 10



Work Step by Step

Since $sec ( x) =\frac{1}{cos(x)}$ and $cos(cos^{-1}x)=x$ The given problem can be solved as: $sec({cos^{-1}(\frac{1}{2})})=\frac{1}{cos(cos^{-1}(\frac{1}{2}))}$ or, $\frac{1}{cos(cos^{-1}(\frac{1}{2}))}$ =$\frac{1}{\frac{1}{2}}$=2 Hence. the final answer is: $2$
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