Answer
$tan^2~x-sec^2~x = -1$
Work Step by Step
$tan^2~x-sec^2~x$
$= \frac{sin^2~x}{cos^2~x} - \frac{1}{cos^2~x}$
$= \frac{sin^2~x-1}{cos^2~x}$
$= \frac{sin^2~x-(sin^2~x+cos^2~x)}{cos^2~x}$
$= \frac{-cos^2~x}{cos^2~x}$
= -1
Trigonometry (10th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
Published by
Pearson
ISBN 10:
0321671775
ISBN 13:
978-0-32167-177-6
Chapter 5 - Test - Page 244: 3
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