If $x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}$ $x = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}$ and $y = \frac{\sqrt{5} - 1}{\sqrt{5} + 1},$ ${\text{y}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}{\text{,}}$ then the value of $\frac{x^{2} + x y + y^{2}}{x^{2} - x y + y^{2}}$ $\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}}$ is?

1 $\frac{3}{4}$ $\frac{3}{4}$
2 $\frac{5}{3}$ $\frac{5}{3}$
3 $\frac{4}{3}$ $\frac{4}{3}$
4 $\frac{3}{5}$ $\frac{3}{5}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} and  y = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \\ ∴ x = \frac{1}{y} \\ \Leftrightarrow x y = 1 \\ x + y = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} + \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \\ \Rightarrow x + y = \frac{5 + 1 + 2 \sqrt{5} + 5 + 1 - 2 \sqrt{5}}{5 - 1} \\ \Rightarrow x + y = \frac{12}{4} \\ \Rightarrow x + y = 3 \\ \Rightarrow x + \frac{1}{x} = 3 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} = {(3)}^{2} - 2 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} = 7 \\ Now, \frac{x^{2} + x y + y^{2}}{x^{2} - x y + y^{2}} \\ = \frac{x^{2} + y^{2} + x y}{x^{2} + y^{2} - x y} \\ = \frac{7 + 1}{7 - 1} \\ = \frac{8}{6} \\ = \frac{4}{3} \end{aligned}

$\eqalign{ & x = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ and y}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \therefore x = \frac{1}{y} \cr & \Leftrightarrow xy = 1 \cr & x + y = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \Rightarrow x + y = \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{5 - 1}} \cr & \Rightarrow x + y = \frac{{12}}{4} \cr & \Rightarrow x + y = 3 \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {\left( 3 \right)^2} - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr & {\text{Now,}}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} \cr & = \frac{{{x^2} + {y^2} + xy}}{{{x^2} + {y^2} - xy}} \cr & = \frac{{7 + 1}}{{7 - 1}} \cr & = \frac{8}{6} \cr & = \frac{4}{3} \cr}$