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:

Given, $$x=\frac{\sqrt{3}}{2}$$

$$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\times\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$$

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

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

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

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

$$=\frac{1+\sqrt{1-\frac{3}{4}}}{\frac{\sqrt{3}}{2}}$$

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

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

$$=\sqrt{3}$$

Hence, the correct answer is Option B