Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Review Exercises - Page 644: 34

Answer

$1$

Work Step by Step

Since, $\sec^2 \theta =1+\tan^2 \theta $ Here, $\theta = \dfrac{\pi}{5}$ Thus, we have $\sec^2 \dfrac{\pi}{5}-\tan^2 \dfrac{\pi}{5}=( \tan^2 \dfrac{\pi}{5}+1)- \tan^2 \dfrac{\pi}{5} $ or, $=1$
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