Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set: 37

Answer

1

Work Step by Step

1 + $tan^{2}x$ = $sec^{2}x$ for all x $sec^{2}x$ - $tan^{2}x$ = 1. So for x = $\frac{\pi}{3}$, $sec^{2}\frac{\pi}{3}$ - $tan^{2}\frac{\pi}{3}$ = 1.
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