Question

Sol. [Key Concept: Take $\tan$ of both sides: $tan(\tan^{- 1}(1 - \alpha) + \tan^{- 1}(1 - \beta)) = 1$; expand and use $\beta = \frac{1}{3\alpha}$ to solve for $\alpha + \beta$.]

$${\ \beta = \frac{1}{3\alpha},\tan^{- 1}(1 - \alpha) + \tan^{- 1}(1 - \beta) = \frac{\pi}{4} }{\Rightarrow tan\left( \tan^{- 1}(1 - \alpha) + \tan^{- 1}(1 - \beta) \right) = 1 }{\Rightarrow \frac{1 - \alpha + 1 - \beta}{1 - (1 - \alpha)(1 - \beta)} = 1 }{\Rightarrow 2 - \alpha - \beta = 1 - (1 - \alpha - \beta + \alpha\beta) }{\Rightarrow 2 - \alpha - \beta = \alpha + \beta - \alpha\beta }{\Rightarrow 2(\alpha + \beta) = 2 + \alpha\beta }{\Rightarrow 2(\alpha + \beta) = 2 + \frac{1}{3} = \frac{7}{3} }{\alpha + \beta = \frac{7}{6} }{6(\alpha + \beta) = 7 }$$