If a = \(\frac{\sqrt {3}}{2}\) then find the value of \(\sqrt {1 + a}\) + \(\sqrt {1 - a}\) + \(\frac{\sqrt {3}}{2}\)

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

a =  \(\frac{\sqrt {3}}{2}\)

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

          = \(\frac{4 + 2\sqrt {3}}{4}\)             (multiply & divide by 2)

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

Similarly,

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

Now,

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

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

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

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

= 3\(\frac{\sqrt {3}}{2}\)