Evaluation of Expression Involving Inverse Tangent Equations

To find the value of $6(\alpha + \beta)$, we begin with the given equation: $\tan^{-1}(1 - \alpha) + \tan^{-1}(1 - \beta) = \frac{\pi}{4}$

Taking the tangent on both sides of the equation and applying the formula $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$, we get: $\frac{(1 - \alpha) + (1 - \beta)}{1 - (1 - \alpha)(1 - \beta)} = \tan\left(\frac{\pi}{4}\right)$

Since $\tan\left(\frac{\pi}{4}\right) = 1$, this simplifies to: $\frac{2 - (\alpha + \beta)}{1 - (1 - \alpha - \beta + \alpha\beta)} = 1$

Simplifying the denominator: $\frac{2 - (\alpha + \beta)}{\alpha + \beta - \alpha\beta} = 1$

Cross-multiplying yields: $2 - (\alpha + \beta) = \alpha + \beta - \alpha\beta$ $2(\alpha + \beta) - \alpha\beta = 2$

We are given that $\beta = \frac{1}{3\alpha}$, which implies: $\alpha\beta = \alpha \cdot \frac{1}{3\alpha} = \frac{1}{3}$

Substituting $\alpha\beta = \frac{1}{3}$ into our simplified equation: $2(\alpha + \beta) - \frac{1}{3} = 2$ $2(\alpha + \beta) = 2 + \frac{1}{3} = \frac{7}{3}$ $\alpha + \beta = \frac{7}{6}$

To verify the given condition $0 < \alpha < 1$: $\alpha + \frac{1}{3\alpha} = \frac{7}{6}$ $6\alpha^2 - 7\alpha + 2 = 0$ $(2\alpha - 1)(3\alpha - 2) = 0$

Thus, $\alpha = \frac{1}{2}$ or $\alpha = \frac{2}{3}$. Both values satisfy $0 < \alpha < 1$.

Now, calculating $6(\alpha + \beta)$: $6(\alpha + \beta) = 6 \left(\frac{7}{6}\right) = 7$

Hence, the correct option is B.