If $$a = \frac{{\sqrt 3 }}{2}{\text{,}}$$   then $$\sqrt {1 + a} + \sqrt {1 - a} = ?$$

Solution(By Examveda Team)

$$\eqalign{ & a = \frac{{\sqrt 3 }}{2}{\text{ (given)}} \cr & \therefore \sqrt {1 + a} + \sqrt {1 - a} \cr & = \sqrt {1 + \frac{{\sqrt 3 }}{2}} + \sqrt {1 - \frac{{\sqrt 3 }}{2}} \cr & = \sqrt {\frac{{2 + \sqrt 3 }}{2}} + \sqrt {\frac{{2 - \sqrt 3 }}{2}} \cr & = \sqrt {\frac{{2\left( {2 + \sqrt 3 } \right)}}{4}} + \sqrt {\frac{{2\left( {2 - \sqrt 3 } \right)}}{4}} \cr & = \sqrt {\frac{{4 + 2\sqrt 3 }}{4}} + \sqrt {\frac{{4 - 2\sqrt 3 }}{4}} \cr} $$
$$ = \sqrt {\frac{{3 + 1 + 2 \times \sqrt 3 \times 1}}{2}} + $$     $$\sqrt {\frac{{3 + 1 - 2 \times \sqrt 3 \times 1}}{2}} $$     \[\because \left\{ \begin{gathered} {\left( {\sqrt 3 } \right)^2} + {\left( 1 \right)^2} - 2.\sqrt 3 .1 = {\left( {\sqrt 3 - 1} \right)^2} \hfill \\ {\left( {\sqrt 3 } \right)^2} + {\left( 1 \right)^2} + 2.\sqrt 3 .1 = {\left( {\sqrt 3 + 1} \right)^2} \hfill \\ {a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2} \hfill \\ {a^2} + {b^2} - 2ab = {\left( {a + b} \right)^2} \hfill \\ \end{gathered} \right\}\]
$$\eqalign{ & = \sqrt {\frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{2}} + \sqrt {\frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{2}} \cr & = \frac{{\sqrt 3 + 1 + \sqrt 3 - 1}}{2} \cr & = \frac{{2\sqrt 3 }}{2} \cr & = \sqrt 3 \cr} $$