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Question Number 10114 by JAZAR last updated on 24/Jan/17

how can demontred tan 3x=_( ) ((3tanx−tan^3 x)/(1_ −3tan^2 x)) pleace help me

$${how}\:{can}\:{demontred} \\ $$$$\mathrm{tan}\:\mathrm{3}{x}\underset{\:\:\:\:\:\:\:\:\:} {=}\frac{\mathrm{3}{tanx}−{tan}^{\mathrm{3}} {x}}{\underset{} {\mathrm{1}}−\mathrm{3}{tan}^{\mathrm{2}} {x}} \\ $$$${pleace}\:{help}\:{me} \\ $$

Answered by mrW1 last updated on 24/Jan/17

tan 3x=tan (x+2x) =((tan x+tan 2x)/(1−tan x tan 2x)) =((tan x+((2tan x)/(1−tan^2 x)))/(1−tan x×((2tan x)/(1−tan^2 x)))) =((tan x (1−tan^2 x)+2tan x)/(1−tan^2 x−2tan^2 x)) =((3tan x−tan^3 x)/(1−3tan^2 x))

$$\mathrm{tan}\:\mathrm{3}{x}=\mathrm{tan}\:\left({x}+\mathrm{2}{x}\right) \\ $$$$=\frac{\mathrm{tan}\:{x}+\mathrm{tan}\:\mathrm{2}{x}}{\mathrm{1}−\mathrm{tan}\:{x}\:\mathrm{tan}\:\mathrm{2}{x}} \\ $$$$=\frac{\mathrm{tan}\:{x}+\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{x}}}{\mathrm{1}−\mathrm{tan}\:{x}×\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{x}}} \\ $$$$=\frac{\mathrm{tan}\:{x}\:\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{x}\right)+\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{x}−\mathrm{2tan}^{\mathrm{2}} \:{x}}\: \\ $$$$=\frac{\mathrm{3tan}\:{x}−\mathrm{tan}^{\mathrm{3}} \:{x}}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} \:{x}} \\ $$

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