$\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} - \dfrac{2 - \sqrt{3}}{2 + \sqrt{3}}$ is equal to:

1. 4

2. $2\sqrt{3}$

3. 1

4. $8\sqrt{3}$

Given, $\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} - \dfrac{2 - \sqrt{3}}{2 + \sqrt{3}}$

Solving,

$\Rightarrow \dfrac{(2 + \sqrt{3}) \times (2 + \sqrt{3})}{(2 - \sqrt{3}) \times (2 + \sqrt{3})} - \dfrac{(2 - \sqrt{3})\times (2 - \sqrt{3})}{(2 + \sqrt{3}) \times (2 - \sqrt{3})}\\[1em] \Rightarrow \dfrac{(2 + \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2} - \dfrac{(2 - \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2}\\[1em] \Rightarrow \dfrac{(2)^2 + (\sqrt{3})^2 + 2 \times 2 \times \sqrt{3}}{4 - 3} - \dfrac{(2)^2 + (\sqrt{3})^2 - 2 \times 2 \times \sqrt{3}}{4 - 3}\\[1em] \Rightarrow \dfrac{4 + 3 + 4\sqrt{3}}{1} - \dfrac{4 + 3 - 4\sqrt{3}}{1}\\[1em] \Rightarrow 7 + 4\sqrt{3} - (7 - 4\sqrt{3})\\[1em] \Rightarrow 7 + 4\sqrt{3} - 7 + 4\sqrt{3}\\[1em] \Rightarrow 8\sqrt{3}.$

Hence, option 4 is correct option.