Question

Find the absolute value of:

$\frac{\sqrt{3}+\frac{1}{\sqrt{3}}}{1-\sqrt{3}}-\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}}$

Answer 1

We have,

$\Rightarrow \frac{\sqrt{3}+\frac{1}{\sqrt{3}}}{1-\sqrt{3}}-\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}}$

$\Rightarrow \frac{3+1}{\sqrt{3}(1-\sqrt{3})}-\frac{3-1}{\sqrt{3}(1+\sqrt{3})}$

$\Rightarrow \frac{4}{\sqrt{3}(1-\sqrt{3})}-\frac{2}{\sqrt{3}(1+\sqrt{3})}$

$\Rightarrow \frac{2}{\sqrt{3}}[\frac{2}{(1-\sqrt{3})}-\frac{1}{(1+\sqrt{3})}]$

$\Rightarrow \frac{2}{\sqrt{3}}\left[\frac{2(1+\sqrt{3})-(1-\sqrt{3})}{(1-3)}\right]$

$\Rightarrow \frac{2}{\sqrt{3}}\left[\frac{2+2\sqrt{3}-1+\sqrt{3}}{-2}\right]$

$\Rightarrow -\frac{1}{\sqrt{3}}\left[1+3\sqrt{3}\right]$

Hence, the answer is 1.