Question Number 68473 by naka3546 last updated on 11/Sep/19
$$\frac{\mathrm{sin}\:\mathrm{72}°}{\mathrm{sin}\:\mathrm{42}°}\:\:=\:\:{p} \\ $$$$\mathrm{tan}\:\mathrm{12}°\:\:=\:\:? \\ $$
Commented by Kunal12588 last updated on 11/Sep/19
$$\frac{{sin}\left(\mathrm{60}°+\mathrm{12}°\right)}{{sin}\left(\mathrm{30}°+\mathrm{12}°\right)}={p} \\ $$$$\Rightarrow\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cos}\mathrm{12}°+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{12}°}{\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{12}°+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sin}\mathrm{12}°}={p} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{3}}+{tan}\mathrm{12}°}{\mathrm{1}+\sqrt{\mathrm{3}}\:{tan}\mathrm{12}°}={p} \\ $$$$\Rightarrow{tan}\:\mathrm{12}°\left(\mathrm{1}−{p}\sqrt{\mathrm{3}}\right)={p}−\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{tan}\:\mathrm{12}\:°=\:\frac{{p}−\sqrt{\mathrm{3}}}{\mathrm{1}−{p}\sqrt{\mathrm{3}}} \\ $$