Let \(0 [JEE Main 2026, 6 Apr (Shift 1)]

Let \(A=\tan^{-1}(1-\alpha)\) and \(B=\tan^{-1}(1-\beta)\).

Given \(A+B=\frac{\pi}{4}\), so \(\tan(A+B)=1\).

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

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

\(2-\alpha-\beta=\alpha+\beta-\alpha\beta\)

\(2=2\alpha+2\beta-\alpha\beta\)

Using \(\beta=\frac{1}{3\alpha}\),

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

Multiplying by \(3\alpha\),

\(6\alpha=6\alpha^2+2-\alpha\)

\(6\alpha^2-7\alpha+2=0\)

\((3\alpha-2)(2\alpha-1)=0\)

\(\alpha=\frac{2}{3}\) or \(\alpha=\frac{1}{2}\)

If \(\alpha=\frac{2}{3}\), then \(\beta=\frac{1}{2}\);

if \(\alpha=\frac{1}{2}\), then \(\beta=\frac{2}{3}\).

So, \(\alpha+\beta=\frac{7}{6}\)

\(6(\alpha+\beta)=6\cdot\frac{7}{6}=7\)