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

$\sqrt{3}$

We know that,

( a + b )² = a² + b² + 2ab

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

Put the value of x in the given equation,

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

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

multiply numerator and denominator by 2 we get,

= $\frac{\sqrt{\frac{4+2\sqrt{3}}{4}}+\sqrt{\frac{4-2\sqrt{3}}{4}}}{\sqrt{\frac{4+2\sqrt{3}}{4}}-\sqrt{\frac{4-2\sqrt{3}}{4}}}$

= $\frac{\sqrt{\frac{4+2\sqrt{3}}{4}}+\sqrt{\frac{4-2\sqrt{3}}{4}}}{\sqrt{\frac{4+2\sqrt{3}}{4}}-\sqrt{\frac{4-2\sqrt{3}}{4}}}$