If a = 3 2 , then 1 + a + 1 - a = ?

Answer: Option D

$$\begin{array}{rl}& a=\frac{\sqrt{3}}{2}\text{ (given)}\\ & \therefore \sqrt{1+a}+\sqrt{1-a}\\ & =\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}\\ & =\sqrt{\frac{2+\sqrt{3}}{2}}+\sqrt{\frac{2-\sqrt{3}}{2}}\\ & =\sqrt{\frac{2\left(2+\sqrt{3}\right)}{4}}+\sqrt{\frac{2\left(2-\sqrt{3}\right)}{4}}\\ & =\sqrt{\frac{4+2\sqrt{3}}{4}}+\sqrt{\frac{4-2\sqrt{3}}{4}}\end{array}$$
$$=\sqrt{\frac{3+1+2\times \sqrt{3}\times 1}{2}}+$$     $$\sqrt{\frac{3+1-2\times \sqrt{3}\times 1}{2}}$$     $$\because \left\{\begin{array}{c}{\left(\sqrt{3}\right)}^2+{\left(1\right)}^2-2.\sqrt{3}.1={\left(\sqrt{3}-1\right)}^2\text{hfill}\\ {\left(\sqrt{3}\right)}^2+{\left(1\right)}^2+2.\sqrt{3}.1={\left(\sqrt{3}+1\right)}^2\text{hfill}\\ a^2+b^2-2ab={\left(a-b\right)}^2\text{hfill}\\ a^2+b^2-2ab={\left(a+b\right)}^2\text{hfill}\end{array}\right\}$$
$$\begin{array}{rl}& =\sqrt{\frac{{\left(\sqrt{3}+1\right)}^2}{2}}+\sqrt{\frac{{\left(\sqrt{3}-1\right)}^2}{2}}\\ & =\frac{\sqrt{3}+1+\sqrt{3}-1}{2}\\ & =\frac{2\sqrt{3}}{2}\\ & =\sqrt{3}\end{array}$$