JEE Main 2025 Apr 7 Shift 1, Mathematics Q8: Trigonometric Ratios And Identities

$x=3\left(\frac{\tan \theta +\sqrt{3}}{1-\sqrt{3}\tan \theta }\right)$

$x(1-\sqrt{3}\tan \theta )=3\tan \theta +3\sqrt{3}$

$\therefore \tan \theta =\frac{x-3\sqrt{3}}{3+\sqrt{3}x}\cdots (1)$

$2\left(\frac{\tan \theta +\frac{1}{\sqrt{3}}}{1-\frac{\tan \theta }{\sqrt{3}}}\right)=y$

$\Rightarrow 2(\sqrt{3}\tan \theta +1)=y(\sqrt{3}-\tan \theta )\cdots (2)$

Using (1) and (2),

$2\left(\dfrac{x-3\sqrt3}{\sqrt3+x}+1\right) = y\left(\sqrt3-\dfrac{x-3\sqrt3}{\sqrt3(\sqrt3+x)}\right)$

$\Rightarrow 2\sqrt3(x-3\sqrt3+x+\sqrt3) = y\left(3(\sqrt3+x)-x+3\sqrt3\right)$

$\Rightarrow 4\sqrt3x-12 = y(2x+6\sqrt3)$

$\Rightarrow xy-2\sqrt3x+3\sqrt3y +6=0$

$\therefore \alpha =-2\sqrt{3},\beta =3\sqrt{3},\gamma =6$

$\alpha^2+\beta^2+\gamma^2 =12+27+36 =\boxed{75}$