## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

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

#### Answer

$2$

#### 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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