If $a(2-\sqrt{3})=b(2+\sqrt{3})=1$ then the value of $a^2-b^2$ is

The correct option is A. $8\sqrt{3}$

Given:

$a(2-\sqrt{3})=b(2+\sqrt{3})=1$

$\Rightarrow a=\frac{1}{(2-\sqrt{3})}$ and $b=\frac{1}{(2+\sqrt{3})}$

We know that, $a^2-b^2=(a+b)(a-b)$ ---(1)

Put the value of $a$ and $b$ in (1)

$(\frac{1}{(2-\sqrt{3})}{)}^2-(\frac{1}{(2+\sqrt{3})}{)}^2$

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

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

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

$=8\sqrt{3}$

$\therefore a^2-b^2=8\sqrt{3}$