Question Number 68352 by mhmd last updated on 09/Sep/19
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{15}°}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{15}°}\:\:\mathrm{is} \\ $$
Answered by $@ty@m123 last updated on 09/Sep/19
$$\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta}=\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\therefore\:{the}\:{given}\:{expression} \\ $$$$=\mathrm{cos}\:\mathrm{30} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$