Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.3 - Right Triangle Trigonometry - Exercise Set - Page 561: 43



Work Step by Step

Solve. $\dfrac{\tan \dfrac{\pi}{3}}{2}-\dfrac{1}{\sec \dfrac{\pi}{6}}$ We have $\dfrac{\tan \dfrac{\pi}{3}}{2}-\dfrac{1}{\sec \dfrac{\pi}{6}}=\dfrac{\sqrt 3}{2}- \dfrac{1}{2/\sqrt 3}$ or, $= \dfrac{\sqrt 3}{2}-\dfrac{\sqrt 3}{2}$ or, $=0$
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