If $a=2+\sqrt{3}$, then find the value of $a-\frac{1}{a}$.

Given, $a=2+\sqrt{3}$

Now let's find the value of $\frac{1}{a}$

So, $\frac{1}{a}=\frac{1}{2+\sqrt{3}}$

Rationalise by multiplying both the numerator and the denominator with $2-\sqrt{3}$

$\frac{1}{a}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$

$=\frac{2-\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}$

$=\frac{2-\sqrt{3}}{2^2-(\sqrt{3}{)}^2}$

$=\frac{2-\sqrt{3}}{4-3}$

$=2-\sqrt{3}$

So, $\frac{1}{a}=2-\sqrt{3}$

Now let's simplify $a-\frac{1}{a}$

$a-\frac{1}{a}$$=2+\sqrt{3}+2-\sqrt{3}$

$=2+2$

$=4$

Hence, the value of $a-\frac{1}{a}$ is 4.