If tanθ1, tanθ2, tanθ3 are the real roots of the x3−(a+1)x2+(b−a)x−b=0 where θ1+θ2+θ3∈(0,π), then θ1+θ2+θ3 is equal to
π/2
π/4
3π/4
π
tanθ1+tanθ2+tanθ3=(a+1)Σtanθ1tanθ2=(b−a) tanθ1tanθ2tanθ3=b∴ tanθ1+θ2+θ3=Σtanθ1−Πtanθ11−Σtanθ1tanθ2 =a+1−b1−(b−a)=1⇒ θ1+θ2+θ3=π4