JEE Main 2026 Apr 6 Shift 1, Mathematics Q13: Inverse Trigonometric Functions

JEE Main2026Apr 6, Shift 1Mathematics

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:

Solution

Take $tan$ of both sides using the addition formula:

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

Cross-multiplying, $2 - α - β = 1 - (1 - α - β + α β) = α + β - α β$ .

Hence $2 (α + β) = 2 + α β$ . With $α β = α \cdot \frac{1}{3 α} = \frac{1}{3}$ :

$2 (α + β) = 2 + \frac{1}{3} = \frac{7}{3} \Rightarrow α + β = \frac{7}{6}$ .

$∴ 6 (α + β) = 7$ .

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