if 3 tan^-1 (1/2+√3) - tan^-1 1/x = tan^-1 1/3 then what is the value of x

since $3{\tan }^{-1}x={\tan }^{-1}\left[\frac{3x-x^3}{1-3x^2}\right]$
$\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$
$$\begin{gathered} 3{\tan }^{-1}\frac{1}{2+\sqrt{3}}=3{\tan }^{-1}\left(2-\sqrt{3}\right) \\ ={\tan }^{-1}\left[\frac{3.(2-\sqrt{3})-(2-\sqrt{3}{)}^3}{1-3.(2-\sqrt{3}{)}^2}\right] \\ ={\tan }^{-1}\left[\frac{(2-\sqrt{3}).\left(3-\{4+3-4\sqrt{3}\}\right)}{1-3.\left(4+3-4\sqrt{3}\right)}\right] \\ ={\tan }^{-1}\left[\frac{(2-\sqrt{3})(-4+4\sqrt{3})}{1-21+12\sqrt{3}}\right] \\ ={\tan }^{-1}\left[\frac{-8+4\sqrt{3}+8\sqrt{3}-12}{-20+12\sqrt{3}}\right] \\ ={\tan }^{-1}\left[\frac{-20+12\sqrt{3}}{-20+12\sqrt{3}}\right]={\tan }^{-1}1 \end{gathered}$$
therefore the given equation can be written as:
$$\begin{gathered} {\tan }^{-1}1-{\tan }^{-1}\frac{1}{x}={\tan }^{-1}\frac{1}{3} \\ {\tan }^{-1}\frac{1}{x}={\tan }^{-1}1-{\tan }^{-1}\frac{1}{3} \\ {\tan }^{-1}\frac{1}{x}={\tan }^{-1}\frac{1-{\displaystyle \frac{1}{3}}}{1+1.{\displaystyle \frac{1}{3}}} [\sin ce {\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\frac{A-B}{1+AB} \end{gathered}$$
$$\begin{gathered} {\tan }^{-1}\frac{1}{x}={\tan }^{-1}\frac{2/3}{4/3} \\ {\tan }^{-1}\frac{1}{x}={\tan }^{-1}\frac{1}{2} \\ \Rightarrow \frac{1}{x}=\frac{1}{2} \\ \Rightarrow x=2 \end{gathered}$$

hope this helps you