Question

若a+b=π/6,则根号3(tanatanb-1)+3(tana+tanb)=

Answer 1

tan(a+b)=(tana+tanb)/(1-tanatanb)

tan(π/6)=(tana+tanb)/(1-tanatanb)

√3/3=(tana+tanb)/(1-tanatanb)

√3(1-tanatanb)=3(tana+tanb)

√3(tanatanb-1)+3(tana+tanb)=0