If $$x=\tfrac{\sqrt3+1}{\sqrt3-1}$$​ and $$y=\tfrac{\sqrt3-1}{\sqrt3+1}$$​, then the value of $$x^2+y^2$$​ is:

Correct option is A

Given:

$$x=\frac{\sqrt3+1}{\sqrt3-1}$$ and  $$y=\frac{\sqrt3-1}{\sqrt3+1}$$

Formula Used:

(a+ b)² = a²+b²+ 2ab

(a - b)² = a²+b²-2ab

Solution:

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

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

=$$\left(\frac{4 + 2\sqrt{3}}{2}\right)^2$$+$$\left(\frac{4 - 2\sqrt{3}}{2}\right)^2$$

=$$\left({2 + \sqrt{3}}\right)^2 +\left({2 - \sqrt{3}}\right)^2$$

=4 + 3+ 2$$\sqrt{3}$$ +4+ 3- 2$$\sqrt3​$$

=7+7 =14

Option (A) is right.​