If $\tan \theta +\tan (\frac{\pi }{3}+\theta )-\tan (\frac{\pi }{3}-\theta )=3,\text{ then }\theta =$

$\tan \theta +\tan \left(\frac{{\displaystyle \pi }}{{\displaystyle 3}}+\theta \right)-\tan \left(\frac{{\displaystyle \pi }}{{\displaystyle 3}}-\theta \right)=3$

$\tan \theta +\frac{\sqrt{3}+\tan \theta }{1-\sqrt{3}\tan \theta }-\frac{\sqrt{3}-\tan \theta }{1+\sqrt{3}\tan \theta }=3$

$\tan \theta (1-3{\tan }^2\theta )+\sqrt{3}+3\tan \theta +\tan \theta +\sqrt{3}{\tan }^2\theta$

$-(\sqrt{3}-\tan \theta -3\tan \theta +\sqrt{3}{\tan }^2\theta )=3(1-3{\tan }^2\theta )$

$\tan \theta -3{\tan }^3\theta +\sqrt{3}+4\tan \theta +\sqrt{3}{\tan }^2\theta -\sqrt{3}$

$+4\tan \theta -\sqrt{3}{\tan }^2\theta =3-9{\tan }^2\theta$

$\tan \theta -3{\tan }^3\theta +8\tan \theta =3-9{\tan }^2\theta$

$\begin{array}{l}\frac{3(3\tan \theta -{\tan }^3\theta )}{1-3{\tan }^2\theta }=3\\ \tan 3\theta =1\end{array}$