If x=frac{sqrt{3}+1}{sqrt{3}-1} and y=frac{sq | vedclass

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(C) Given: $x = \frac{\sqrt{3}+1}{\sqrt{3}-1}$ and $y = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.Rationalizing $x$:$x = \frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3

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$14$

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(C) Given: $x = \frac{\sqrt{3}+1}{\sqrt{3}-1}$ and $y = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.Rationalizing $x$:$x = \frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{3+1+2\sqrt{3}}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$.Rationalizing $y$:$y = \frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3+1-2\sqrt{3}}{3-1} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$.Now,calculate $x^{2}+y^{2}$:$x^{2}+y^{2} = (2+\sqrt{3})^{2} + (2-\sqrt{3})^{2}$.Using the identity $(a+b)^{2} + (a-b)^{2} = 2(a^{2}+b^{2})$:$x^{2}+y^{2} = 2(2^{2} + (\sqrt{3})^{2}) = 2(4+3) = 2(7) = 14$.

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

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