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:

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Inverse Trigonometric Functions

Let

0 < α < 1, β = \frac{1}{3 α}

and

tan^{- 1} (1 - α) + tan^{- 1} (1 - β) = \frac{π}{4}

. Then

6 (α + β)

is equal to:

Answer: (B)

With

β = \frac{1}{3 α}

, so

α β = \frac{1}{3}

, take the tangent of

tan^{- 1} (1 - α) + tan^{- 1} (1 - β) = \frac{π}{4}

\frac{( 1 - α ) + ( 1 - β )}{1 - ( 1 - α ) ( 1 - β )} = 1

. This simplifies to

2 - α - β = α + β - α β

, i.e.

2 (α + β) = 2 + α β = 2 + \frac{1}{3} = \frac{7}{3}

, so

α + β = \frac{7}{6}

. Therefore

6 (α + β) = 7

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