Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 693: 35


$\displaystyle \frac{3\tan x-\tan^{3}x}{1-\tan^{3}x}$

Work Step by Step

$\displaystyle \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$ $ \displaystyle \tan 2A=\frac{2\tan A}{1-\tan^{2}A}$ ----------------- $\tan 3x=\tan(x+2x)$ $=\displaystyle \frac{\tan x+\tan 2x}{1-\tan x\tan 2x}$ $=\displaystyle \frac{\tan x+\frac{2\tan x}{1-\tan^{2}x}}{1-\tan x\cdot\frac{2\tan x}{1-\tan^{2}x}}=\frac{\frac{\tan x(1-\tan^{2}x)+2\tan x}{1-\tan^{2}x}}{\frac{(1-\tan^{2}x)-2\tan^{2}x}{1-\tan^{2}x}}$ $=\displaystyle \frac{\tan x(1-\tan^{2}x)+2\tan x}{(1-\tan^{2}x)-2\tan^{2}x}$ $=\displaystyle \frac{3\tan x-\tan^{3}x}{1-\tan^{3}x}$
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