JEE Main 2026 — Mathematics Trigonometry

Given $\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \dfrac{\pi}{4}$tan−1(1−α)+tan−1(1−β)=4π​

Applying the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$tan−1x+tan−1y=tan−1(1−xyx+y​), we get:

$$\dfrac{(1-\alpha) + (1-\beta)}{1 - (1-\alpha)(1-\beta)} = \tan\left(\dfrac{\pi}{4}\right) = 1$$1−(1−α)(1−β)(1−α)+(1−β)​=tan(4π​)=1

$$2 - \alpha - \beta = 1 - (1 - \alpha - \beta + \alpha\beta)$$2−α−β=1−(1−α−β+αβ)

$$2 - \alpha - \beta = \alpha + \beta - \alpha\beta$$2−α−β=α+β−αβ

$$2(\alpha + \beta) = 2 + \alpha\beta$$2(α+β)=2+αβ

Given $\beta = \dfrac{1}{3\alpha}$β=3α1​, we have $\alpha\beta = \dfrac{1}{3}$αβ=31​. Substituting this into the equation:

$$2(\alpha + \beta) = 2 + \dfrac{1}{3}$$2(α+β)=2+31​

$$2(\alpha + \beta) = \dfrac{7}{3}$$2(α+β)=37​

$$\alpha + \beta = \dfrac{7}{6}$$α+β=67​

Therefore, $6(\alpha + \beta) = 6 \times \dfrac{7}{6} = 7$6(α+β)=6×67​=7

Answer: $7$7