Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 418: 81



Work Step by Step

Since $\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}}$ Then, $\dfrac{\sin{40^{\circ}}}{\cos{40^{\circ}}} = \tan{40^{\circ}}$ Therefore, $\tan{40^{\circ}} - \dfrac{\sin{40^{\circ}}}{\cos{40^{\circ}}} = \tan{40^{\circ}}-\tan{40^{\circ}} = \boxed{0}$
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