The maximum value of $\mathrm{f}\left(\mathrm{x}\right)={\tan }^{-1} \left(\frac{(\sqrt{12}-2){\mathrm{x}}^2}{{\mathrm{x}}^4+2{\mathrm{x}}^2+3}\right)$ is

$\begin{array}{rr}\mathrm{f}\left(\mathrm{x}\right)={\tan }^{-1}\left(\frac{(\sqrt{12}-2){\mathrm{x}}^2}{{\mathrm{x}}^4+2{\mathrm{x}}^2+3}\right)& \\ ={\tan }^{-1}\left(\frac{(\sqrt{4\times 3}-2){\mathrm{x}}^2}{{\mathrm{x}}^2({\mathrm{x}}^2+2+{\displaystyle \frac{3}{{\mathrm{x}}^2})}}\right)& \\ ={\tan }^{-1}\left(\frac{(2\sqrt{3}-2)}{{\mathrm{x}}^2+\frac{3}{{\mathrm{x}}^2}+2}\right)& \\ \mathrm{f}\left(\mathrm{x}\right)={\tan }^{-1}\left(\frac{2(\sqrt{3}-1)}{{\mathrm{x}}^2+\frac{3}{{\mathrm{x}}^2}+2}\right)& \\ \text{ As }{\mathrm{x}}^2+\frac{3}{{\mathrm{x}}^2}\ge 2\sqrt{3}[\mathrm{usin}\mathrm{g}\text{ A.M. }\ge \mathrm{G}\cdot \mathrm{M}]& \\ \Rightarrow {\mathrm{x}}^2+\frac{3}{{\mathrm{x}}^3}+2\ge 2+2\sqrt{3}(\because adding 2 on both sides)& \\ now \left(\frac{2(\sqrt{3}-1)}{{\mathrm{x}}^2+\frac{3}{{\mathrm{x}}^2}+2}\right)\ge \left(\frac{2(\sqrt{3}-1)}{2+2\sqrt{3}}\right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}& \\ \therefore \left(\mathrm{f}\right(\mathrm{x}){)}_{\max }={\tan }^{-1}\left(\frac{2(\sqrt{3}-1)}{2(\sqrt{3}+1)}\right)=\frac{\mathrm{\pi }}{12}& \mathrm{ }\end{array}$