Answer
$-1$
Work Step by Step
Use the identity $tan(x)=\frac{sin(x)}{cos(x)}$ and $sec(x)=\frac{1}{cos(x)}$, we have $tan^2x-sec^2x=(\frac{sin(x)}{cos(x)})^2-(\frac{1}{cos(x)})^2=\frac{sin^2x-1}{cos^2x}=\frac{-cos^2x}{cos^2x}=-1$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
Published by
Pearson
ISBN 10:
013421742X
ISBN 13:
978-0-13421-742-0
Chapter 7 - Trigonometric Identities and Equations - Test - Page 740: 3
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