Question Number 42345 by soufiane zarik last updated on 23/Aug/18
$$\mathrm{tan}\:\mathrm{15}°\:= \\ $$
Commented by MJS last updated on 24/Aug/18
$$\mathrm{tan}\:\mathrm{15}°\:=\mathrm{tan}\:\frac{\mathrm{30}°}{\mathrm{2}}\:=\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{1}+\mathrm{cos}\:\mathrm{30}°}=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}=\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$
Commented by soufiane zarik last updated on 24/Aug/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by $@ty@m last updated on 25/Aug/18
$${Alternative}; \\ $$$$\mathrm{tan}\:\left(\mathrm{45}−\mathrm{30}\right)=\frac{\mathrm{tan}\:\mathrm{45}−\mathrm{tan}\:\mathrm{30}}{\mathrm{1}+\mathrm{tan}\:\mathrm{45}.\mathrm{tan}\:\mathrm{30}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\:=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}+\mathrm{1}} \\ $$