If tanθ1, tan⁡θ2, tan⁡θ3 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

tan⁡θ1+tan⁡θ2+tan⁡θ3=(a+1)Σtan⁡θ1tan⁡θ2=(b−a) tan⁡θ1tan⁡θ2tan⁡θ3=b∴ tan⁡θ1+θ2+θ3=Σtan⁡θ1−Πtan⁡θ11−Σtan⁡θ1tan⁡θ2 =a+1−b1−(b−a)=1⇒ θ1+θ2+θ3=π4