If x=$$\frac{\sqrt{3}+1}{\sqrt{3}-1}$$  y=$$\frac{\sqrt{3}-1}{\sqrt{3}+1}$$ then find the value of  $$x^{2}+y^{2}$$

$$x=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\left(\sqrt{3}+1\right)^2}{3-1}=\frac{1}{2}\left(3+2\sqrt{3}+1\right)=\left(2+\sqrt{3}\right).$$

Similarly, $$y=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\left(2-\sqrt{3}\right).$$

So, $$x^2=\left(2+\sqrt{3}\right)^2=4+3+4\sqrt{3}=\left(7+4\sqrt{3}\right).$$

And, $$y^2=\left(2-\sqrt{3}\right)^2=4+3-4\sqrt{3}=\left(7-4\sqrt{3}\right).$$

So, $$x^2+y^2=7+4\sqrt{3}+7-4\sqrt{3}=14.$$

A is correct choice.