If x = 3 2 , then 1 + x 1 + 1 + x + 1 - x 1 - 1 - x is equal to?

Answer: Option B

$$\begin{array}{rl}& x=\frac{\sqrt{3}}{2}\\ & \text{or }1+x=1+\frac{\sqrt{3}}{2}\\ & \Rightarrow 1+x=\frac{2+\sqrt{3}}{2}\\ & \Rightarrow 1+x=\frac{2\left(2+\sqrt{3}\right)}{2\times 2}\\ & \left(\text{Divided and multiply by 2}\right)\\ & \Rightarrow 1+x=\frac{4+2\sqrt{3}}{4}\\ & \Rightarrow 1+x=\frac{1+3+2\sqrt{3}}{4}\\ & \Rightarrow 1+x=\frac{4+2\sqrt{3}}{4}\\ & \Rightarrow 1+x=\frac{{\left(1\right)}^2+{\left(\sqrt{3}\right)}^2+\mathrm{2.1.}\sqrt{3}}{4}\\ & \Rightarrow 1+x=\frac{{\left(1+\sqrt{3}\right)}^2}{4}\\ & \therefore \sqrt{1+x}=\frac{1+\sqrt{3}}{2}\\ & \\ & \mathbf{Similarly}:\\ & \sqrt{1-x}\\ & =\frac{\sqrt{3}-1}{2}\\ & \therefore \frac{\sqrt{1+x}}{1+\sqrt{1+x}}\text{ + }\frac{\sqrt{1-x}}{1-\sqrt{1-x}}\\ & =\frac{\frac{1+\sqrt{3}}{2}}{1+\frac{1+\sqrt{3}}{2}}+\frac{\frac{\sqrt{3}-1}{2}}{1-\frac{\sqrt{3}-1}{2}}\\ & =\frac{1+\sqrt{3}}{3+\sqrt{3}}+\frac{\sqrt{3}-1}{3-\sqrt{3}}\\ & =\frac{1+\sqrt{3}}{\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{1-\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-1\right)}\\ & =\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}\\ & =\frac{2}{\sqrt{3}}\end{array}$$