Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX D - Trigonometry - D Exercises: 40

Answer

(a) $tan(x+y)=\frac{tanx+tany}{1-tanxtany}$ (b)$tan(x-y)=\frac{tanx-tany}{1+tanxtany}$

Work Step by Step

(a) $tan(x+y)=\frac{sin(x+y)}{cos(x+y)}$ $sin(x+y)=sinxcosy+cosxsiny$ and $cos(x+y)=cosxcosy-sinxsiny$ Thus, $tan(x+y)=\frac{sinxcosy+cosxsiny}{cosxcosy-sinxsiny}$ Divide the numerator and denominator by $cosx$ $cosy$: $tan(x+y)=\frac{\frac{sinx}{cosx}+\frac{siny}{cosy}}{1-\frac{sinx siny}{cosxcosy}}$ Hence, $tan(x+y)=\frac{tanx+tany}{1-tanxtany}$ (b) $tan(x-y)=\frac{sin(x-y)}{cos(x-y)}$ $sin(x-y)=sinxcosy-cosxsiny$ and $cos(x-y)=cosxcosy+sinxsiny$ Thus, $tan(x-y)=\frac{sinxcosy-cosxsiny}{cosxcosy+sinxsiny}$ Divide the numerator and denominator by $cosx$ $cosy$: $tan(x-y)=\frac{\frac{sinx}{cosx}-\frac{siny}{cosy}}{1+\frac{sinx siny}{cosxcosy}}$ Hence, $tan(x-y)=\frac{tanx-tany}{1+tanxtany}$
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