Question

How do you simplify $\frac{\sqrt{3}-1}{1+\sqrt{3}}$?

Answer 1

Simplify the expression:

We will simplify the given expression $\frac{\sqrt{3}-1}{1+\sqrt{3}}$ by rationalising the denominator.

To rationalise the denominator, multiplying both numerator and denominator of given term $\frac{\sqrt{3}-1}{1+\sqrt{3}}$ by $\sqrt{3}-1$.

$$\begin{gathered} \frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{\sqrt{3}-1}{1+\sqrt{3}}·\frac{\sqrt{3}-1}{\sqrt{3}-1} \\ \Rightarrow \frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{{\left(\sqrt{3}-1\right)}^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)} \\ \Rightarrow \frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{{\left(\sqrt{3}\right)}^2-2\sqrt{3}+1^2}{{\left(\sqrt{3}\right)}^2-1^2}\left[\because \left(a-b\right)\left(a+b\right)=a^2-b^2;{\left(a-b\right)}^2=a^2-2ab+b^2\right] \\ \Rightarrow \frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{3-2\sqrt{3}+1}{3-1} \\ \Rightarrow \frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{4-2\sqrt{3}}{2} \\ \Rightarrow \frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{2} \\ \therefore \frac{\sqrt{3}-1}{1+\sqrt{3}}=2-\sqrt{3} \end{gathered}$$

Hence, $\frac{\sqrt{3}-1}{1+\sqrt{3}}=2-\sqrt{3}$.