If a = \(\frac{\sqrt {5}}{3}\) then find the value of \(\sqrt {1 + a}\) + \(\sqrt {1 - a}\) + \(\frac{\sqrt {5}}{3}\)?

\(\sqrt{5}\) \(\frac{6 \;+ \; \sqrt{6}}{3\sqrt{6}}\)

a =  \(\frac{\sqrt {5}}{3}\)

1 + a = 1 + \(\frac{\sqrt {5}}{3}\) = \(\frac{3 + \sqrt {5}}{3}\)

          = \(\frac{6 + 2\sqrt {3}}{6}\)             (multiply & divide by 2)

          = \(\frac{(\sqrt {5} + 1)^2}{6}\)

Similarly,

1 - a = \(\frac{(\sqrt {5} - 1)^2}{6}\)

Now,

⇒ \(\sqrt {1 + a}\) + \(\sqrt {1 - a}\) + \(\frac{\sqrt {5}}{3}\)  = \(\sqrt {(\frac{\sqrt {3} + 1)^2}{4}}\) + \(\sqrt {(\frac{\sqrt {3} - 1)^2}{4}}\) + \(\frac{\sqrt {3}}{2}\)

= \(\frac{\sqrt {5}\;+\;1}{\sqrt{6}}\) + \(\frac{\sqrt {5}\;-\;1}{\sqrt{6}}\) +  \(\frac{\sqrt {5}}{3}\)

= \(\frac{\sqrt {5}\;+\;1\;+\sqrt {3}\;-\;1}{\sqrt{6}}\) +  \(\frac{\sqrt {5}}{3}\)

= \(\frac{2\sqrt{5}}{\sqrt{6}}\) + \(\frac{\sqrt {5}}{3}\)

=\(\sqrt{5}\) \(\frac{6 \;+ \; \sqrt{6}}{3\sqrt{6}}\)