Answer
$\sqrt{2}$
Work Step by Step
$\sec{(\theta+ 2 \pi k}) = \sec{\theta}$
\geq$\therefore \sec{\dfrac{17 \pi}{4}} = \sec{\left(\dfrac{\pi}{4} + (2 \pi \times 2) \right)} = \sec{\dfrac{\pi}{4}}$
$\sec{\dfrac{\pi}{4}} = \boxed{\sqrt{2}}$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 417: 23
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