Question Number 189428 by Rupesh123 last updated on 16/Mar/23
Answered by BaliramKumar last updated on 16/Mar/23
![tanx tan(60° + x) tan(60° − x) tanx[((tan60° + tanx)/(1−tan60° tanx))][((tan60° − tanx)/(1 + tan60° tanx))] tanx[(((√3) + tanx)/(1−(√3) tanx))][(((√3) − tanx)/(1 + (√3) tanx))] tanx[(((3 − tan^2 x)/(1−3 tan^2 x))] [(((3tanx − tan^3 x)/(1−3 tan^2 x))] = tan3x](https://www.tinkutara.com/question/Q189433.png)
$${tanx}\:{tan}\left(\mathrm{60}°\:+\:{x}\right)\:{tan}\left(\mathrm{60}°\:−\:{x}\right)\: \\ $$$${tanx}\left[\frac{{tan}\mathrm{60}°\:+\:{tanx}}{\mathrm{1}−{tan}\mathrm{60}°\:{tanx}}\right]\left[\frac{{tan}\mathrm{60}°\:−\:{tanx}}{\mathrm{1}\:+\:{tan}\mathrm{60}°\:{tanx}}\right] \\ $$$${tanx}\left[\frac{\sqrt{\mathrm{3}}\:+\:{tanx}}{\mathrm{1}−\sqrt{\mathrm{3}}\:{tanx}}\right]\left[\frac{\sqrt{\mathrm{3}}\:−\:{tanx}}{\mathrm{1}\:+\:\sqrt{\mathrm{3}}\:{tanx}}\right] \\ $$$${tanx}\left[\frac{\left(\mathrm{3}\:−\:{tan}^{\mathrm{2}} {x}\right.}{\mathrm{1}−\mathrm{3}\:{tan}^{\mathrm{2}} {x}}\right] \\ $$$$\left[\frac{\left(\mathrm{3}{tanx}\:−\:{tan}^{\mathrm{3}} {x}\right.}{\mathrm{1}−\mathrm{3}\:{tan}^{\mathrm{2}} {x}}\right]\:=\:{tan}\mathrm{3}{x} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 18/Mar/23
Excellent!