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If for θ ∈ [- π 3, 0 ], the points (x, y)= (3 tan (θ+ π 3 ), 2 tan (θ+ π 6 ) ) lie on x y+α x+β y+γ=0, then α^2+β^2+γ^2 is equal to :

JEE Main 2025 — Mathematics Trigonometry

2025mcqmedium

If for θ∈[−π3,0]\theta \in\left[-\frac{\pi}{3}, 0\right]θ∈[−3π​,0], the points (x,y)=(3tan⁡(θ+π3),2tan⁡(θ+π6))(x, y)=\left(3 \tan \left(\theta+\frac{\pi}{3}\right), 2 \tan \left(\theta+\frac{\pi}{6}\right)\right)(x,y)=(3tan(θ+3π​),2tan(θ+6π​)) lie on xy+αx+βy+γ=0x y+\alpha x+\beta y+\gamma=0xy+αx+βy+γ=0, then α2+β2+γ2\alpha^2+\beta^2+\gamma^2α2+β2+γ2 is equal to :

Official previous-year question

Held on 7 Apr 2025 · Verified 6 Jul 2026.

Options

  1. A

    80

  2. B

    72

  3. C

    96

  4. D

    75

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