Question

If $x=2+\sqrt{3},xy=1$, find $\frac{x}{2-x}+\frac{y}{2-y}$.

• A. $-\sqrt{3}$
• B. $\sqrt{3}$
• C. $-\sqrt{2}$
• D. $-2$

Answer 1

The correct option is D $-2$

given that

$x=2+\sqrt{3}$ and $xy=1$

we can find value of $y$

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

we have to find the value of given expression

$\frac{x}{2-x}+\frac{y}{2-y}$

$=\frac{x\times (2-y)+y\times (2-x)}{(2-x)(2-y)}$

$=\frac{2x-xy+2y-xy}{(2-x)(2-y)}$

$=\frac{2x+2y-2xy}{(2-x)(2-y)}$

$=2\times \frac{x+y-xy}{[4-2(x+y)+xy]}$

now putting the values

$=2\times \frac{2+\sqrt{3}+2-\sqrt{3}-1}{[4-2(2+\sqrt{3}+2-\sqrt{3})+1]}$

$=2\times \frac{3}{[4-8+1]}$

$=-2$