若0＜α＜π/2,π＜β＜3π/2,且tanα=1/7,tanβ=3/4,则α+β=?

若0＜α＜π/2,π＜β＜3π/2,且tanα=1/7,tanβ=3/4,则α+β=?
若0＜α＜π/2,π＜β＜3π/2,且tanα=1/7,tanβ=3/4,则α+β=?

若0＜α＜π/2,π＜β＜3π/2,且tanα=1/7,tanβ=3/4,则α+β=?
tanα=1/7,tanβ=3/4
tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)
=(1/7+3/4)/(1-3/4*1/7)
=1
又0＜α＜π/2,π＜β＜3π/2
所以π＜α+β＜2π
所以α+β=5π/4

0＜α＜π/2,π＜β＜3π/2
π＜α+β＜2π
tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=1,
∴α+β=5π/4