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: 56


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}$
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