$\mathrm{If} 3{\tan }^{-1}\left(\frac{1}{2+\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)={\tan }^{-1}\left(\frac{1}{2}\right)$ then   2021x is equal to

Given equation can be written as
$\begin{array}{l}3{\tan }^{-1}(2-\sqrt{3})-{\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)={\tan }^{-1}\left(\frac{1}{2}\right)\\ \Rightarrow \frac{3\mathrm{\pi }}{12}-{\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)={\tan }^{-1}\left(\frac{1}{2}\right)\\ \Rightarrow {\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)=\frac{\mathrm{\pi }}{4}-{\tan }^{-1}\left(\frac{1}{2}\right)\end{array}$
Taking tangent on both sides, we get
$\frac{1}{\mathrm{x}}=\frac{1-\frac{1}{2}}{1+\frac{1}{2}}\Rightarrow \mathrm{x}=3$