设α=arctan3∈（0,π/2）β=arctan2∈（0,π/2）∴tanα=3,tanβ=2∴tan(α-β)=(tanα-tanβ)/(1+tanαtanβ)=(3-2)/(1+2*3）=1/7∵α-β∈(0,π)∴α-β=arctan1/7∴arctan3-arctan2=arctan1/7