If $$x=\frac{(√3+1)}{(√3-1)}$$​ and $$y=\frac{(√3-1)}{(√3+1)}$$​, then find the value of $$x^2+y^2.$$​​

Correct option is C

​Given:

​$$x = \frac{\sqrt{3}+1}{\sqrt{3}-1}, \quad y = \frac{\sqrt{3}-1}{\sqrt{3}+1}$$​​

Formula Used:

​$$x \times y = 1, \quad x^2 + y^2 = (x+y)^2 - 2xy$$​​

Solution:
Simplify x and y: 

​$$x = \frac{\sqrt{3}+1}{\sqrt{3}-1} \\ \ \\ x = \frac{\sqrt{3}+1}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1} \\ \ \\ =2 + \sqrt3$$​​

​$$y = \frac{\sqrt{3}-1}{\sqrt{3}+1} \\ \ \\ y = \frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}- 1}{\sqrt{3}-1} \\ \ \\ = 2 - \sqrt3$$​​

​Sum: $$x+y = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$$​​

Product: $$x \cdot y = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$$​​

Now:

​$$x^2 + y^2 = 4^2 - 2(1) = 16 - 2 = 14$$