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(B) Given the equation $	an^{-1}(1 - \alpha) + 	an^{-1}(1 - \beta) = \frac{\pi}{4}$.Using the formula $	an^{-1}x + 	an^{-1}y = 	an^{-1}\left(\fra

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$7$

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(B) Given the equation $	an^{-1}(1 - \alpha) + 	an^{-1}(1 - \beta) = \frac{\pi}{4}$.Using the formula $	an^{-1}x + 	an^{-1}y = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$, we get $\frac{(1-\alpha)+(1-\beta)}{1-(1-\alpha)(1-\beta)} = 	an\left(\frac{\pi}{4}ight) = 1$.This simplifies to $2 - (\alpha + \beta) = 1 - (1 - \alpha - \beta + \alpha\beta)$.$2 - \alpha - \beta = \alpha + \beta - \alpha\beta \Rightarrow 2 = 2(\alpha + \beta) - \alpha\beta$.Substitute $\beta = \frac{1}{3\alpha}$ into the equation:$2 = 2(\alpha + \frac{1}{3\alpha}) - \alpha(\frac{1}{3\alpha}) = 2\alpha + \frac{2}{3\alpha} - \frac{1}{3}$.$2 + \frac{1}{3} = 2\alpha + \frac{2}{3\alpha} \Rightarrow \frac{7}{3} = \frac{6\alpha^2 + 2}{3\alpha}$.$7\alpha = 6\alpha^2 + 2 \Rightarrow 6\alpha^2 - 7\alpha + 2 = 0$.Factoring the quadratic: $(2\alpha - 1)(3\alpha - 2) = 0$.Thus, $\alpha = \frac{1}{2}$ or $\alpha = \frac{2}{3}$.If $\alpha = \frac{1}{2}$, then $\beta = \frac{1}{3(1/2)} = \frac{2}{3}$.If $\alpha = \frac{2}{3}$, then $\beta = \frac{1}{3(2/3)} = \frac{1}{2}$.In both cases, $\alpha + \beta = \frac{1}{2} + \frac{2}{3} = \frac{7}{6}$.Therefore, $6(\alpha + \beta) = 6 	imes \frac{7}{6} = 7$.

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:

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