Question Number 84136 by mathocean1 last updated on 09/Mar/20
$${show}\:{that}: \\ $$$${tan}\mathrm{3}{x}=\frac{\mathrm{3}+{tan}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {x}}×{tanx} \\ $$
Answered by Rio Michael last updated on 09/Mar/20
$$\mathrm{tan3}{x}\:=\:\mathrm{tan}\left(\mathrm{2}{x}\:+\:{x}\right)\:=\:\frac{\mathrm{tan2}{x}\:+\:\mathrm{tan}{x}}{\mathrm{1}−\mathrm{tan2}{x}\mathrm{tan}{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\frac{\mathrm{2tan}{x}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}}\:+\:\mathrm{tan}{x}}{\mathrm{1}−\left(\frac{\mathrm{2tan}{x}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}}\right)\mathrm{tan}{x}}\:=\:\frac{\mathrm{2tan}{x}\:+\:\mathrm{tan}{x}\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}\right)}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}−\mathrm{2tan}^{\mathrm{2}} {x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{tan}{x}\:\left(\mathrm{2}\:+\:\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}\right)}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} {x}}\:=\:\frac{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} {x}}\:\mathrm{tan}{x} \\ $$
Commented by mathocean1 last updated on 09/Mar/20
$${thank}\:{you}\:{sir}! \\ $$