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: 44

Answer

$-\dfrac {1}{2}\left( 5\sqrt {2}+4\sqrt {3}\right) $

Work Step by Step

$\tan 285=\tan \left( 225+60\right) =\dfrac {\tan 225+\tan 60}{1-\tan 225\tan 60}=\dfrac {\dfrac {\sqrt {2}}{2}+\sqrt {3}}{1-\dfrac {\sqrt {2}}{2}\sqrt {3}}=\dfrac {\sqrt {2}+\sqrt {3}}{1-\sqrt {6}}$ $=\dfrac {\left( \sqrt {2}+\sqrt {3}\right) \left( 2+\sqrt {6}\right) }{\left( 2-\sqrt {6}\right) \left( 2+\sqrt {6}\right) }=\dfrac {2\sqrt {2}+2\sqrt {3}+2\sqrt {3}+3\sqrt {2}}{2^{2}-\left( \sqrt {6}\right) ^{2}}=\dfrac {5\sqrt {2}+4\sqrt {3}}{-2}=-\dfrac {1}{2}\left( 5\sqrt {2}+4\sqrt {3}\right) $
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