If $$x = \frac{{\sqrt 3 }}{2}{\text{,}}$$   then the value of $$\left( {\frac{{\sqrt {1 + x} + \sqrt {1 - x} }}{{\sqrt {1 + x} - \sqrt {1 - x} }}} \right)\,{\text{is}} = ?$$

$$\eqalign{ & x = \frac{{\sqrt 3 }}{2} \cr & = \frac{{\sqrt {1 + x} + \sqrt {1 - x} }}{{\sqrt {1 + x} - \sqrt {1 - x} }} \times \frac{{\sqrt {1 + x} + \sqrt {1 - x} }}{{\sqrt {1 + x} + \sqrt {1 - x} }} \cr & = \frac{{{{\left( {\sqrt {1 + x} + \sqrt {1 - x} } \right)}^2}}}{{{{\left( {\sqrt {1 + x} } \right)}^2} - {{\left( {\sqrt {1 - x} } \right)}^2}}} \cr & = \frac{{1 + x + 1 - x + 2\sqrt {1 - {x^2}} }}{{1 + x - 1 + x}} \cr & = \frac{{2 + 2\sqrt {1 - {x^2}} }}{{2x}} \cr & = \frac{{1 + \sqrt {1 - {x^2}} }}{x} \cr & = \frac{{1 + \sqrt {1 - \frac{3}{4}} }}{{\sqrt 3 }} \times 2 \cr & = \frac{{\left( {1 + \frac{1}{2}} \right)}}{{\sqrt 3 }} \times 2 \cr & = \frac{{\frac{3}{2} \times 2}}{{\sqrt 3 }} \cr & = \sqrt 3 \cr} $$