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Question Number 84136 by mathocean1 last updated on 09/Mar/20

show that: tan3x=((3+tan^2 x)/(1−3tan^2 x))×tanx

$${show}\:{that}: \\ $$$${tan}\mathrm{3}{x}=\frac{\mathrm{3}+{tan}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {x}}×{tanx} \\ $$

Answered by Rio Michael last updated on 09/Mar/20

tan3x = tan(2x + x) = ((tan2x + tanx)/(1−tan2xtanx)) = ((((2tanx)/(1−tan^2 x)) + tanx)/(1−(((2tanx)/(1−tan^2 x)))tanx)) = ((2tanx + tanx(1−tan^2 x))/(1−tan^2 x−2tan^2 x)) = ((tanx (2 + 1−tan^2 x))/(1−3tan^2 x)) = ((3−tan^2 x)/(1−3tan^2 x)) tanx

$$\mathrm{tan3}{x}\:=\:\mathrm{tan}\left(\mathrm{2}{x}\:+\:{x}\right)\:=\:\frac{\mathrm{tan2}{x}\:+\:\mathrm{tan}{x}}{\mathrm{1}−\mathrm{tan2}{x}\mathrm{tan}{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\frac{\mathrm{2tan}{x}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}}\:+\:\mathrm{tan}{x}}{\mathrm{1}−\left(\frac{\mathrm{2tan}{x}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}}\right)\mathrm{tan}{x}}\:=\:\frac{\mathrm{2tan}{x}\:+\:\mathrm{tan}{x}\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}\right)}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}−\mathrm{2tan}^{\mathrm{2}} {x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{tan}{x}\:\left(\mathrm{2}\:+\:\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}\right)}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} {x}}\:=\:\frac{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} {x}}\:\mathrm{tan}{x} \\ $$

Commented by mathocean1 last updated on 09/Mar/20

thank you sir!

$${thank}\:{you}\:{sir}! \\ $$

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