Question

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

Answer 1

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

Answer 2

tan(a+b)=(tana+tanb)/(1-tana*tanb)=1
π<a+b<2π,
a+b=5π/4