Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises - Page 738: 103


$\dfrac {294+125\sqrt {6}}{92}$

Work Step by Step

$\tan \left( \arcsin \dfrac {3}{5}+\arccos \dfrac {5}{7}\right) =\dfrac {\tan \left( \arcsin \dfrac {3}{5}\right) +\tan \left( \arccos \dfrac {5}{7}\right) }{1-\tan \left( \arcsin \dfrac {3}{5}\right) \tan \left( \arccos \dfrac {5}{7}\right) }=\dfrac {\dfrac {\dfrac {3}{5}}{\sqrt {1-\left( \dfrac {3}{5}\right) ^{2}}}+\dfrac {\sqrt {1-\left( \dfrac {5}{7}\right) ^{2}}}{\dfrac {5}{7}}}{1-\dfrac {\dfrac {3}{5}}{\sqrt {1-\left( \dfrac {3}{5}\right) ^{2}}}\dfrac {\sqrt {1-\left( \dfrac {5}{7}\right) ^{2}}}{\dfrac {5}{7}}}$ $=\dfrac {\dfrac {3}{4}+\dfrac {2\sqrt {6}}{5}}{1-\dfrac {3}{4}\times \dfrac {2\sqrt {6}}{5}}=\dfrac {\dfrac {15+8\sqrt {6}}{20}}{1-\dfrac {6\sqrt {6}}{20}}=\dfrac {15+8\sqrt {6}}{20-6\sqrt {6}}=\dfrac {\left( 15+8\sqrt {6}\right) \left( 20+6\sqrt {6}\right) }{\left( 20-6\sqrt {6}\right) \left( 20+6\sqrt {6}\right) }=\dfrac {300+288+250\sqrt {6}}{184}=\dfrac {294+125\sqrt {6}}{92}$
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