Question

若实数ab满足(a+1)(b+1)=2,则arctana+arctanb等于多少

Answer 1

解:

设α=arctana,β=arctanb

α、β∈(0,π/2)

则tanα=a,tanβ=b

∵(a+1)(b+1)=2

∴a+b=1-ab

于是

tan(α+β)=(a+b)/(1-ab)=1

∴α+β=π/4

即arctana+arctanb=π/4