If x = \(\sqrt {1 + \frac{\sqrt {3}}{2}}\) - \(\sqrt {1 - \frac{\sqrt {3}}{2}}\),

then find value of \(\frac{\sqrt {2}\;-\;x}{\sqrt {2}\;+\;x}\).

3 - 2\(\sqrt {2}\)

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

Multiply and divide by 2 for making perfect square

x = \(\sqrt {1+\frac{2\sqrt {3}}{4}}\) - \(\sqrt {1-\frac{2\sqrt {3}}{4}}\)

x = \(\sqrt {\frac{{(\sqrt {3}+1)}^{2}}{4}}\) - \(\sqrt {\frac{{(\sqrt {3}-1)}^{2}}{4}}\)

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

x = 1

Put & find

⇒ \(\frac{\sqrt {2}\;-\;x}{\sqrt {2}\;+\;x}\) = \(\frac{\sqrt {2}\;-\;1}{\sqrt {2}\;+\;1}\) = \(\frac{\sqrt {2}\;-\;1}{\sqrt {2}\;+\;1}\) × \(\frac{\sqrt {2}\;-\;1}{\sqrt {2}\;-\;1}\) = (\(\sqrt {2}\) - 1)² = 3 - 2\(\sqrt {2}\)