Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Section 1.3 - Trigonometric Functions - Exercises 1.3 - Page 28: 55

Answer

See the step-by-step proof below.

Work Step by Step

Recall $ sin(A+B)=sinAcosB+sinBcosA $ and $ cos(A+B)=cosAcosB-sinAsinB $ We have: $ tan(A+B)=\frac{sin(A+B)}{cos(A+B)}=\frac{sinAcosB+sinBcosA}{cosAcosB-sinAsinB}$ Divide both the numerator and the denominator by $ cosAcosB $: $ tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$ which, is the formula.
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