Let  $\frac{\tan 8^0}{1-3{\tan }^{2}8}==\frac{1}{8}[\frac{8\tan 8^0}{1-3{\tan }^{2}8}+\tan 8^0-\tan 8^0]$

$\frac{\tan 8^0}{1-3{\tan }^{2}8}==\frac{3}{8}\left(\frac{3\tan 8^0-{\tan }^{3}8}{1-3{\tan }^{2}8}\right)-\frac{1}{8}\left(\tan 8^0\right)$

$\frac{\tan 8°}{1-3{\tan }^{2}8°}=-\frac{1}{8}\tan 8^0+\frac{3}{8}\tan {24}^0$

$\frac{3\tan {24}^0}{1-3{\tan }^{2}24}=\frac{-3}{8}\tan 24°+\frac{9}{8}\tan 72°$

$\frac{9\tan {72}^0}{1-3{\tan }^{2}72}=-\frac{9}{8}\tan {72}^0+\frac{27}{8}\tan {216}^0$

$\frac{27\tan {216}^0}{1-3{\tan }^{2}216}=-\frac{27}{8}\tan {216}^0+\frac{81}{8}\tan {648}^0$

$$\begin{gathered} \frac{\tan 8^0}{1-3{\tan }^{2}8}+\frac{3\tan {24}^0}{1-3{\tan }^{2}24}+\frac{9\tan {72}^0}{1-3{\tan }^{2}72}+\frac{27\tan {216}^0}{1-3{\tan }^{2}216} \\ =\frac{81}{8}\tan {648}^0-\frac{1}{8}\tan 8^0 \end{gathered}$$

$\tan 648°=\tan (3\times 180°+108°)=\tan 108°$
$\therefore$ Solving LHS
 $\frac{81}{8}\tan {108}^0-\frac{1}{8}\tan 8^0$