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: 30



Work Step by Step

Since, $\sec^2 \theta =1+\tan^2 \theta $ Here, $\theta = 1.4^{\circ}$ We have: $ \tan^2 1.4^{\circ}-\sec^2 1.4^{\circ} $ or, $= \tan^2 1.4^{\circ}-(1+\tan^2 1.4^{\circ} ) $ or, $=-1$
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