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

$$\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \quad (\text{for } xy < 1)$$

$$\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \frac{\pi}{4}$$

$$\frac{(1-\alpha) + (1-\beta)}{1 - (1-\alpha)(1-\beta)} = \tan\left(\frac{\pi}{4}\right)$$

$$\frac{2 - \alpha - \beta}{1 - (1 - \alpha - \beta + \alpha\beta)} = 1$$

$$\frac{2 - \alpha - \beta}{\alpha + \beta - \alpha\beta} = 1$$

$$2 - \alpha - \beta = \alpha + \beta - \alpha\beta$$

$$2 - 2(\alpha + \beta) + \alpha\beta = 0 \quad \text{--- (Equation 1)}$$

$$\beta = \frac{1}{3\alpha} \implies \alpha\beta = \frac{1}{3}$$

$$2 - 2(\alpha + \beta) + \frac{1}{3} = 0$$

$$\frac{7}{3} = 2(\alpha + \beta)$$

$$6(\alpha + \beta) = 3 \times \left[2(\alpha + \beta)\right]$$

$$6(\alpha + \beta) = 3 \times \frac{7}{3} = 7$$