Find the exact value of trigonometric expression: $ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$? How can I simplify this trigonometric expression?

$$  \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$$

I used 
$$\sin \frac{\pi}{15}=2 \sin \frac{\pi}{30}  \cos \frac{\pi}{30}$$ and $$\cos\frac{\pi}{15}=2\cos^2\frac{\pi}{30}-1$$
But these give me more complicated expressions.

$\begin{align}
T &= \large \left(\frac{2\cos 6° \cos 12°}{1\;+\;2\cos 6° \sin 12°}\right)
\left({2\sin 6° \over 2\sin 6°}\right)\cr 
&= \large \frac{\sin 24°}{2\sin 6°+\;2\sin^2 12°}\cr\cr
{1\over T}&= \large \frac{2\sin(30°-24°)+\;(1-\cos 24°)}{\sin 24°}\cr
&=\large {(\cos 24° - \sqrt3\sin24°) + (1-\cos 24°) \over \sin 24°}\cr
&=\large{1\over\sin(60°-36°)} - \sqrt3 \cr
&=\large{4\over 2\sqrt3 \cos36° - 2\sin 36°}-\sqrt3 \cr
\end{align}$
From wikipedia on Golden triangle, 
$\quad2\cos(36°)=\phi,\quad 2\sin(36°)=\sqrt{4-\phi^2}$ 
$\begin{align}
\large{1\over T}&=\left({4\over \sqrt3\phi-\sqrt{4-\phi^2}}\right)
\left({\sqrt3\phi+\sqrt{4-\phi^2} \over \sqrt3\phi+\sqrt{4-\phi^2}} \right)
-\sqrt3 \cr
&= {4(\sqrt3\phi + \sqrt{4-\phi^2}) \over 3\phi^2-(4-\phi^2)} -\sqrt3 \cr
&= {\sqrt3\phi + \sqrt{4-\phi^2} \over \phi^2-1}-\sqrt3 \cr
&= {\sqrt3\phi + \sqrt{4-\phi^2} \over \phi} -\sqrt3 \cr
&= {\sqrt{4-\phi^2} \over \phi} = \tan(36°)
\end{align}$
$$T = \tan(90°-36°) = \tan(54°)$$
