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 - Page 679: 34



Work Step by Step

$\tan \alpha =\dfrac {1}{\cot \alpha }=\dfrac {1}{\tan \left( 90-\alpha \right) }\Rightarrow \tan 24=\dfrac {1}{\cot 24}=\dfrac {1}{\tan 66}$
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