Find the value of $\frac{\sin 24x}{\cos18x \cos 6x}$. Given  $\tan x = \sqrt{5} - 2 $, find the value of $$\frac{\sin 24x}{\cos 18x\cos 6x}.$$
Now I wrote numerator as $\sin (18 + 6)x$ and I used identity $\sin(A+B)$, after that I reduced problem to $\tan 18x + \tan 6x$
How do I proceed?
Thanks

We have
\begin{align}
\tan 2x&=\frac{2\tan x}{1-\tan^2x}\\
&=\frac{2(\sqrt{5}-2)}{1-(\sqrt{5}-2)^2}\\
&=\frac{2(\sqrt{5}-2)}{1-5+4\sqrt{5}-4}\\
&=\frac{2(\sqrt{5}-2)}{4(\sqrt{5}-2)}\\
&=\frac{1}{2}
\end{align}
Then use
$$\tan 3A=\frac{3\tan A-\tan^3A}{1-3\tan^2A}$$
to deduce that 
$$\tan 6x=\frac{11}{2}$$
Use it once again to deduce that 
$$\tan18x=\frac{1199}{718}$$
