Mathematics - Trigonometric Equation Question with Solution | TestHub

$\begin{array}{rl}& x=3\left(\frac{\tan \theta +\sqrt{3}}{1-\sqrt{3}\tan \theta }\right)\\ & x-\sqrt{3}\tan \theta =3\tan \theta +3\sqrt{3}\\ & \tan \theta =\frac{x-3\sqrt{3}}{3+\sqrt{3}x}\dots (1)\\ & 2\left(\frac{\tan \theta +\frac{1}{\sqrt{3}}}{1-\frac{\tan \theta }{\sqrt{3}}}=y\right)\\ & 2(\sqrt{3}\tan \theta +1)=y(\sqrt{3}-\tan \theta )\dots .\end{array}$
using (1) and (2)
$\begin{array}{rl}& 2\left(\frac{x-3\sqrt{3}}{\sqrt{3}+x}+1\right)=y\left(\sqrt{3}-\frac{(x-3\sqrt{3})}{\sqrt{3}(\sqrt{3}+x)}\right)\\ & 2\sqrt{3}(x-3\sqrt{3}+x+\sqrt{3})=y(3(\sqrt{3}+x)-x+3\sqrt{3})\\ & 4\sqrt{3}x-12=y(2x+6\sqrt{3})\\ & xy-2\sqrt{3}x+3\sqrt{3}y-6=0\\ & \Rightarrow \alpha =-2\sqrt{3},\beta =3\sqrt{3},\gamma =-6\\ & {\alpha }^2+{\beta }^2+{\gamma }^2=12+27+36=75\end{array}$