$\frac{\tan8°}{1-3\tan^2 8°}+\frac{3\tan24°}{1-3\tan^2 24°} + \frac{9\tan72°}{1-3\tan^2 72°} + \frac{27\tan216°}{1-3\tan^2 216°}=x\tan108°+y\tan8°$ $$\frac {\tan 8°}{1-3\tan^2 8°}+\frac {3\tan 24°}{1-3\tan^2 24°} + \frac{9\tan 72°}{1-3\tan^2 72°} + \frac{27\tan 216°}{1-3\tan^2 216°} =x\tan108°+y\tan8° .$$
Find the value of $x$ and $y$.
I tried different approaches like using $$ \tan 3x=\frac{3\tan x-\tan^3 x}{1-3\tan^2 x}$$
and converting into $\sin$ and $\cos$, but I was unable to simplify.
Can anybody help?

$$\frac {\tan 8°}{1-3\tan^2 8°}+\frac {3\tan 24°}{1-3\tan^2 24°} + \frac{9\tan 72°}{1-3\tan^2 72°} + \frac{27\tan 216°}{1-3\tan^2 216°} =x\tan108°+y\tan8°$$
Now, $$\frac {\tan 8°}{1-3\tan^2 8°}=\frac{1}{8}\left(\frac {8 \tan 8°}{(1-3\tan^2 8°)}+\tan 8°-\tan8° \right) \\ 
=\frac{1}{8}\left(-\tan8° \right)+\frac{1}{8}\left(\frac {8 \tan 8°}{(1-3\tan^2 8°)}+\tan 8° \right) \\ 
=\frac{1}{8}\left(-\tan8° \right)+\frac{3}{8}(\tan 24°)$$
Solving in this way, we get the LHS as
$$\frac{81}{8}\tan(108^\circ)-\frac{1}{8}\tan(8^\circ)$$
From this, we get $$x+y=10$$
