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