Precalculus (6th Edition) Blitzer

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

Chapter 5 - Review Exercises - Page 708: 40


$-\frac{\sqrt 3}{3}$

Work Step by Step

Using the double-angle formula $tan2\theta=\frac{2tan\theta}{1-tan^2\theta}$, we have $\frac{2tan\frac{5\pi}{12}}{1-tan^2\frac{5\pi}{12}}=tan2(\frac{5\pi}{12})=tan\frac{5\pi}{6}=-\frac{\sqrt 3}{3}$
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