Let 0<α<1, β=1/(3 α) and tan^-1(1-α)+ tan^-1(1-β)=(π)/4. Then 6(α+β) is equal to:

Let $\displaystyle 0<\alpha<1, \beta=\frac{1}{3 \alpha}$ and $\displaystyle \tan ^{-1}(1-\alpha)+\tan ^{-1}(1-\beta)=\frac{\pi}{4}$. Then $\displaystyle 6(\alpha+\beta)$ is equal to: