If tan⁡ θ1,tan ⁡θ2,tan⁡ θ3are the real roots of the x3−(a+1)x2 +(b−a)x−b=0, where θ1+θ2+θ3∈(0,π),then θ1+θ2+θ3is 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