If $x=\frac{\sqrt{3}}{2}$ , then the value of  $\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$  is $\frac{k}{\sqrt{3}}$ , then $k=$

On rationalizing

$\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{{\left(\sqrt{1+x}-\sqrt{1-x}\right)}^2}{{\left(\sqrt{1+x}\right)}^2-{\left(\sqrt{1-x}\right)}^2}$

$=\frac{2-2\sqrt{1-x^2}}{2x}$

$ifx=\frac{\sqrt{3}}{2}$ then

$=\frac{2\left(1-\sqrt{1-\frac{3}{4}}\right)}{2\times \frac{\sqrt{3}}{2}}$

$=\frac{1}{2}\times \frac{2}{\sqrt{3}}$

$=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{k}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\therefore k=1$