Let 0 < α < 1, β = 1/(3α) and tan⁻¹(1-α) + tan⁻¹(1-β) = π/4. Then 6(α + β) is equal to:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q13 – JEE-MAIN 2026

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Home Maths Inverse TrigonometryQuestion #232

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:

Clue 1 of 3: Inverse Trigonometric Identity

Recall the sum formula for inverse tangent functions.

🧠 Full Solution Path

Step 1: Apply Inverse Tangent Sum Formula✦ Active

Given the equation $\tan^{- 1} (1 - α) + \tan^{- 1} (1 - β) = \frac{π}{4}$ . Since $0 < α < 1$ , $1 - α \in (0, 1)$ . Also, $β = \frac{1}{3 α}$ , so $1 - β$ can be positive or negative. In all valid cases, the condition for the standard formula $\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})$ (i.e., $x y < 1$ ) holds. Applying the formula:

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

\frac{2 - (α + β)}{1 - (1 - α - β + α β)} = \tan (\frac{π}{4})

\frac{2 - (α + β)}{α + β - α β} = 1

2 - (α + β) = α + β - α β

2 = 2 (α + β) - α β

Step 2: Substitute and Form Quadratic Equation○ Expand

Substitute the given relation $β = \frac{1}{3 α}$ into the simplified equation from Step 1:

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

2 = 2 α + \frac{2}{3 α} - \frac{1}{3}

Multiply the entire equation by $3 α$ to eliminate denominators and rearrange into a quadratic equation:

6 α = 6 α^{2} + 2 - α

6 α^{2} - 7 α + 2 = 0

Step 3: Solve for

α

and Calculate Final Expression○ Expand

Solve the quadratic equation $6 α^{2} - 7 α + 2 = 0$ for $α$ using the quadratic formula:

α = \frac{- (- 7) \pm \sqrt{(- 7)^{2} - 4 (6) (2)}}{2 (6)} = \frac{7 \pm \sqrt{49 - 48}}{12} = \frac{7 \pm 1}{12}

This yields two possible values for $α$ : $α_{1} = \frac{7 + 1}{12} = \frac{8}{12} = \frac{2}{3}$ and $α_{2} = \frac{7 - 1}{12} = \frac{6}{12} = \frac{1}{2}$ . Both values satisfy the condition $0 < α < 1$ . Let's calculate $α + β$ for each case:

Case 1: If $α = \frac{2}{3}$ , then $β = \frac{1}{3 (2 / 3)} = \frac{1}{2}$ . So $α + β = \frac{2}{3} + \frac{1}{2} = \frac{4 + 3}{6} = \frac{7}{6}$ .

Case 2: If $α = \frac{1}{2}$ , then $β = \frac{1}{3 (1 / 2)} = \frac{2}{3}$ . So $α + β = \frac{1}{2} + \frac{2}{3} = \frac{3 + 4}{6} = \frac{7}{6}$ .

In both valid cases, $α + β = \frac{7}{6}$ . Finally, calculate $6 (α + β)$ :

6 (α + β) = 6 (\frac{7}{6}) = 7

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Maths Question 13 – JEE-MAIN 2026

Clue 1 of 3: Inverse Trigonometric Identity

🧠 Full Solution Path