Question

已知tana,tanβ是方程x²+(3根号3)x+4=0的两根,且a,β∈(-π/2，π/2),则a+β=

Answer 1

tana+tanb=-3根号3
tana*tanb=4
楼上的式子借下了哈：tan（a+b）=-3根号3/(1-4)=根号3
a+b=π/3,因为tan（π/3)=根号3

Answer 2

用韦达定理求tan(a+b)=(tana+tanb)/[1-tana*tanb}
a,β∈(-π/2，π/2)