Question

tan15°=tan(45°-30°)

Answer 1

tanα－tanβ tan（α－β）＝—————— 1＋tanα ·tanβ tan15°=tan(45°-30°)=（1-√3/3)/(1+√3/3) =(1-√3/3)^2/(2/3) =(3+1-2√3)/2=2--√3