Precalculus (10th Edition)

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

Chapter 7 - Analytic Trigonometry - 7.5 Sum and Difference Formulas - 7.5 Assess Your Understanding - Page 487: 30



Work Step by Step

Recall that: $\tan(\alpha-\beta)=\dfrac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$ Hence, $\dfrac{\tan(40^\circ)-\tan(10^\circ)}{1+\tan(40^\circ)\tan(10^\circ)}\\ =\tan(40^\circ-10^\circ)\\ =\tan(30^\circ)\\ =\dfrac{\sqrt3}{3}$
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