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: 43

Answer

$\sqrt {3}-2$

Work Step by Step

$\tan165=\tan \left( 120+45\right) =\dfrac {\tan 120+\tan 45}{1-\tan {120}\tan 45}=\dfrac {-\sqrt {3}+1}{1-\left( -\sqrt {3}\right) \times 1}=\dfrac {1-\sqrt {3}}{1+\sqrt {3}}=$ $\dfrac {\left( 1-\sqrt {3}\right) ^{2}}{\left( 1+\sqrt {3}\right) \left( 1-\sqrt {3}\right) }=\dfrac {\left( 1-\sqrt {3}\right) ^{2}}{1^{2}-\sqrt {3}^{2}}=\dfrac {\left( 1-2\sqrt {3}+3\right) }{-2}=\sqrt {3}-2$
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