Question

If x = $\frac{\sqrt{3}}{2}$ show that $\frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}=1$

Answer 1

Disclaimer: In the given question, the RHS of the equation is given wrong .Instead of 2 it will be 1

$\mathrm{LHS}=\frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}$

$$\begin{gathered} \mathrm{We}\mathrm{have}1+x=1+\frac{\sqrt{3}}{2}=\frac{2+\sqrt{3}}{2} \\ \sqrt{1+x}=\sqrt{1+\frac{\sqrt{3}}{2}}=\sqrt{\frac{3}{4}+\frac{1}{4}+\frac{\sqrt{3}}{2}} \\ =\sqrt{{\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2+2\times \frac{\sqrt{3}}{2}\times \frac{1}{2}} \\ =\sqrt{{\left(\frac{\sqrt{3}}{2}+\frac{1}{2}\right)}^2} \\ =\frac{\sqrt{3}+1}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{and}1-x=1-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{2} \\ \sqrt{1-x}=\sqrt{1-\frac{\sqrt{3}}{2}}=\sqrt{\frac{3}{4}+\frac{1}{4}-\frac{\sqrt{3}}{2}} \\ =\sqrt{{\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2-2\times \frac{\sqrt{3}}{2}\times \frac{1}{2}} \\ =\sqrt{{\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)}^2} \\ =\frac{\sqrt{3}-1}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{in}\mathrm{LHS},\mathrm{we}\mathrm{get}: \\ \frac{{\displaystyle \frac{2+\sqrt{3}}{2}}}{1+{\displaystyle \frac{\sqrt{3}+1}{2}}}+\frac{{\displaystyle \frac{2-\sqrt{3}}{2}}}{1-{\displaystyle \frac{\sqrt{3}-1}{2}}} \\ =\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1} \\ =\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}} \end{gathered}$$

$$\begin{gathered} =\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left[\left(3\right)+\left(\sqrt{3}\right)\right]\left[\left(3\right)-\left(\sqrt{3}\right)\right]} \\ =\frac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{{\left(3\right)}^2-{\left(\sqrt{3}\right)}^2} \\ =\frac{6}{6} \\ =1 \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{LHS}=\mathrm{RHS}=1 \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$