Question Number 143499 by mnjuly1970 last updated on 15/Jun/21
Commented by MJS_new last updated on 15/Jun/21
$$ \\ $$$$\mathrm{tan}\:\left(\mathrm{60}°−\alpha\right)\:=\frac{\sqrt{\mathrm{3}}−\mathrm{tan}\:\alpha}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\:\alpha} \\ $$$$\mathrm{tan}\:\left(\mathrm{60}°+\alpha\right)\:=\frac{\sqrt{\mathrm{3}}+\mathrm{tan}\:\alpha}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{tan}\:\alpha} \\ $$$$\mathrm{tan}\:\mathrm{3}\alpha\:=\frac{\left(\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:\alpha\right)\mathrm{tan}\:\alpha}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} \:\alpha} \\ $$$$\mathrm{now}\:\mathrm{simply}\:\mathrm{let}\:\mathrm{tan}\:\alpha\:={t} \\ $$
Answered by justtry last updated on 15/Jun/21
