sinα=4/5,α属于(π/2,π)
∴cosα=-√(1-sin²α)=-3/5
∴tanα=sinα/cosα=-4/3
于是
tan(α+β)
=(tanα+tanβ)/(1-tanαtanβ)
=(-4/3+1/2)/(1+4/3x1/2)
=(-5/6)/(10/6)
=-1/2