Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.2 Trigonometric Functions: Unit Circle Approach - 6.2 Assess Your Understanding - Page 379: 86

Answer

$\frac{2\sqrt3}{3}$.

Work Step by Step

$\tan{60^\circ}+\tan{150^\circ}=\tan{\frac{\pi}{3}}+\tan{\frac{5\pi}{6}}=\sqrt3-\frac{\sqrt3}{3}=\frac{2\sqrt3}{3}$.
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