Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.3 Sum and Difference Identities - 7.3 Exercises: 54

Answer

$2+\sqrt{3}$

Work Step by Step

Use the subtraction formula for tangent, $\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$. $\tan\left(-\frac{7\pi}{12}\right)$ $=\tan\left(-\frac{\pi}{4}-\frac{\pi}{3}\right)$ $=\frac{\tan(-\frac{\pi}{4})-\tan\frac{\pi}{3}}{1+\tan(-\frac{\pi}{4})\tan\frac{\pi}{3}}$ $=\frac{-1-\sqrt{3}}{1+(-1)*\sqrt{3}}$ $=\frac{-1-\sqrt{3}}{1-\sqrt{3}}$ $=\frac{-1-\sqrt{3}}{1-\sqrt{3}}*\frac{1+\sqrt{3}}{1+\sqrt{3}}$ $=\frac{-1-2\sqrt{3}-3}{1-3}$ $=\frac{-4-2\sqrt{3}}{-2}$ $=2+\sqrt{3}$
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