$$\theta = {\tan ^{ - 1}}\left( {2\,{{\tan }^2}\theta } \right) - {\tan ^{ - 1}}\left( {\frac{1}{3}\tan \theta } \right){\text{ then }}\tan \theta = $$

Solution :
$$\eqalign{ & \theta = {\tan ^{ - 1}}\left[ {\frac{{2\,{{\tan }^2}\theta - \frac{1}{3}\tan \theta }}{{1 + \frac{2}{3}{{\tan }^3}\theta }}} \right] \cr & \Rightarrow \tan \theta = \frac{{6\,{{\tan }^2}\theta - \tan \theta }}{{3 + 2\,{{\tan }^3}\theta }} \cr & \Rightarrow 1 = \frac{{6\tan \theta - 1}}{{3 + 2\,{{\tan }^3}\theta }}{\text{ or }}\tan \theta = 0 \cr & \Rightarrow 2\,{\tan ^3}\theta - 6\tan \theta + 4 = 0 \cr & \Rightarrow {\left( {\tan \theta - 1} \right)^2}\left( {\tan \theta + 2} \right) = 0 \cr & \Rightarrow \tan \theta = 1;\tan \theta = - 2;\tan \theta = 0. \cr} $$