If $$\tan \alpha = \sqrt{3} + 2,$$ then the value of $$\tan \alpha - \cot \alpha$$ is

$$\tan \alpha = \sqrt{3} + 2$$    Eq.(1)

$$\tan \alpha - \cot \alpha$$

$$\tan \alpha - \frac{1}{\tan \alpha}$$

$$\frac{\tan \alpha^{2} - {1}} {\tan \alpha}$$    Eq.(2)

Put Eq.(1) in Eq.(2).

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

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

$$\frac{6+4\sqrt{3}}{\sqrt{3}+2}$$

Now multiply $$(\sqrt{3} - 2)$$ in numerator and denominator.

$$\frac{6+4\sqrt{3}}{\sqrt{3}+2}$$

$$(\frac{6+4\sqrt{3}}{\sqrt{3}+2})\times(\frac{\sqrt{3}-2}{\sqrt{3}-2})$$

After apply multiplication got $$\frac{6\sqrt{3}-12+12-8\sqrt{3}}{-1}$$

$$\frac{-2\sqrt{3}}{-1}$$

$${2\sqrt{3}}$$