设 arctan1/2=α,arctan1/3=β;
则有,tanα=1/2,tanβ=1/3;
所以
tan(arctan1/2+arctan1/3)=tan(α+β)=(tanα+tanβ)/(1-tanα·tanβ)
=(1/2+1/3)/(1 - 1/2 X 1/3) = 1
设 arctan1/2=α,arctan1/3=β;
则有,tanα=1/2,tanβ=1/3;
所以
tan(arctan1/2+arctan1/3)=tan(α+β)=(tanα+tanβ)/(1-tanα·tanβ)
=(1/2+1/3)/(1 - 1/2 X 1/3) = 1