tanα=1/2,tanβ=1/3,α、β均为锐角,则α+β=

tanα=1/2,tanβ=1/3,α、β均为锐角,则α+β=
tanα=1/2,tanβ=1/3,α、β均为锐角,则α+β=

tanα=1/2,tanβ=1/3,α、β均为锐角,则α+β=
tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)
=(1/2+1/3)/(1-1/6)
=1
又因为α、β均为锐角,所以 0

tan（α+β)=(tanα+tanβ)/(1-tanαtanβ)=[(1/2)+(1/3)]/[1-(1/2)*(1/3)]=(5/6)/(1-1/6)=1.又α、β为锐角，所以0＜α+β＜π。于是α+β=π/4.

tan(a+b)=(tana+tanb)/(1-tana*tanb)
=(1/2+1/3)/[1-(1/2)*(1/3)]
=1
所以，α+β=π/4