$\theta ={\tan }^{-1} \left(2{\tan }^2 \theta \right)-{\tan }^{-1} \left(\right(1/3)\tan \theta )\text{ if }$tan $\theta$ is equal to

$\theta ={\tan }^{-1} \left(2{\tan }^2 \theta \right)-{\tan }^{-1} \left(\right(1/3)\tan \theta )$

$\begin{array}{r}\Rightarrow \tan \theta =\frac{2{\tan }^2 \theta -(1/3)\tan \theta }{1+(2/3){\tan }^3 \theta }\\ \Rightarrow \tan \theta \frac{2\tan \theta -(1/3)}{1+(2/3){\tan }^3 \theta }-1=0\end{array}$

which is true if tan $\theta$ = 0

or

 $$\begin{gathered} \frac{2\mathrm{can} \theta -(1/3)}{1+(2/3){\tan }^2 \theta }=1 \\ \Rightarrow {\tan }^3 \theta -3\tan \theta +2=0 \end{gathered}$$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\tan \theta -1{)}^2(\tan \theta +2)=0\end{array}$ which holds if $\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\tan \theta =1$ or $\tan \theta =-2$