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, $=1$
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13446-914-3
ISBN 13:
978-0-13446-914-0
Chapter 4 - Review Exercises - Page 644: 34
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