Given expression
\( \Large \frac{1 + x}{1+\sqrt{1+x}} + \frac{1-x}{1-\sqrt{1-x}} \)
On putting \( \Large x = \frac{\sqrt{3}}{2} \), we get
\( \Large \frac{1 + \frac{\sqrt{3}}{2}}{1+\sqrt{1 + \frac{\sqrt{3}}2}} + \frac{1 - \frac{\sqrt{3}}{2}}{1 - \sqrt{1-\frac{\sqrt{3}}2{}}} \)
= \( \Large \frac{2 + \sqrt{3}}{\sqrt{2} \left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right) } + \frac{2 - \sqrt{3}}{\sqrt{2} \left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right) } \)
= \( \Large \frac{2 + \sqrt{3}}{2 + \sqrt{4 + 2\sqrt{3}}} + \frac{2 - \sqrt{3}}{2 - \sqrt{4-2\sqrt{3}}} \)
=> \( \Large \frac{2 + \sqrt{3}}{2 + \sqrt{ \left(1\right)^{2} + \left(\sqrt{3}\right)^{2} } + 2\sqrt{3}} + \frac{2 - \sqrt{3}}{2 - \sqrt{ \left(1\right)^{2} + \left(\sqrt{3}\right)^{2} - 2\sqrt{3} }} \)
=> \( \Large \frac{2 + \sqrt{3}}{2 + \sqrt{ \left(1 + \sqrt{3}\right) ^{2}}} + \frac{2 - \sqrt{3}}{2 - \sqrt{ \left(\sqrt{3} - 1\right)^{2} }} \)
= \( \Large \frac{2 + \sqrt{3}}{2 + \left(1 + \sqrt{3}\right) } + \frac{2 - \sqrt{3}}{2 - \left(\sqrt{3} - 1\right) } \)
= \( \Large \frac{2+\sqrt{3}}{3+\sqrt{3}} + \frac{2-\sqrt{3}}{3-\sqrt{3}} \)
= \( \Large \frac{ \left(2+\sqrt{3}\right) \left(3 - \sqrt{3}\right) + \left(2 - \sqrt{3}\right) \left(3 + \sqrt{3}\right) }{ \left(3 + \sqrt{3}\right) \left(3 - \sqrt{3}\right) } \)
=> \( \Large \frac{6+3\sqrt{3}-2\sqrt{3}-3+6-3\sqrt{3}+2\sqrt{3}-3}{ \left(3 + \sqrt{3}\right) \left(3 - \sqrt{3}\right) } \)
=> \( \Large \frac{6}{9 + 3\sqrt{3} - 3\sqrt{3} - 3} = \frac{6}{6} = 1 \)
