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.
 A: $\begin{align}
T &= \large \left(\frac{2\cos 6° \cos 12°}{1\;+\;2\cos 6° \sin 12°}\right)
\left({2\sin 6° \over 2\sin 6°}\right)\cr 
&= \large \frac{\sin 24°}{2\sin 6°+\;2\sin^2 12°}\cr\cr
{1\over T}&= \large \frac{2\sin(30°-24°)+\;(1-\cos 24°)}{\sin 24°}\cr
&=\large {(\cos 24° - \sqrt3\sin24°) + (1-\cos 24°) \over \sin 24°}\cr
&=\large{1\over\sin(60°-36°)} - \sqrt3 \cr
&=\large{4\over 2\sqrt3 \cos36° - 2\sin 36°}-\sqrt3 \cr
\end{align}$
From wikipedia on Golden triangle, 
$\quad2\cos(36°)=\phi,\quad 2\sin(36°)=\sqrt{4-\phi^2}$ 
$\begin{align}
\large{1\over T}&=\left({4\over \sqrt3\phi-\sqrt{4-\phi^2}}\right)
\left({\sqrt3\phi+\sqrt{4-\phi^2} \over \sqrt3\phi+\sqrt{4-\phi^2}} \right)
-\sqrt3 \cr
&= {4(\sqrt3\phi + \sqrt{4-\phi^2}) \over 3\phi^2-(4-\phi^2)} -\sqrt3 \cr
&= {\sqrt3\phi + \sqrt{4-\phi^2} \over \phi^2-1}-\sqrt3 \cr
&= {\sqrt3\phi + \sqrt{4-\phi^2} \over \phi} -\sqrt3 \cr
&= {\sqrt{4-\phi^2} \over \phi} = \tan(36°)
\end{align}$
$$T = \tan(90°-36°) = \tan(54°)$$
A: Let $\arctan\left(\dfrac{2\cos6^\circ\cos12^\circ}{1+2\cos6^\circ\sin12^\circ}\right)=y,-90^\circ<y<90^\circ$
$\implies\dfrac{2\cos6^\circ\cos12^\circ}{1+2\cos6^\circ\sin12^\circ}=\tan y=\dfrac{\sin y}{\cos y}$
Rearranging we get $$\cos(6^\circ+y)+\cos(18^\circ+y)=\sin y=\cos(90^\circ-y)$$
$$\iff\cos(90^\circ-y)=\cos(18^\circ+y)+\cos(6^\circ+y)=2\cos6^\circ\cos(12^\circ+y)\ \ \ \  (1)$$
Like Solve equation $\tan(x)=\sec(42^\circ)+\sqrt{3}$
using Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
$$\cos36^\circ=\cos72^\circ+\cos60^\circ=2\cos6^\circ\cos66^\circ\ \ \ \  (2)$$
Compare $(1),(2)$
$$\frac{\cos36^\circ}{\cos(90^\circ-y)}=\dfrac{2\cos6^\circ\cos66^\circ}{2\cos6^\circ\cos(12^\circ+y)}$$
$$\iff\dfrac{\cos(12^\circ+y)}{\cos(90^\circ-y)}=\dfrac{\cos66^\circ}{\cos36^\circ}$$
Apply Componendo et Dividendo and Prosthaphaeresis Formulas to find 
$$\tan(y-39^\circ)=\tan15^\circ$$
The rest should be easy!
A: The identity can also be proved from: $\quad\cos36° = \cos 72° + {1\over2}$ 
$\cos 72° + {1\over2}= (2\cos^2 36° - 1)+ {1\over2}\;= \large{\phi^2\over2}-{1\over2} ={\phi \over 2}=\cos36°$

$T= \Large \frac{2\cos 6° \cos 12°}{1\;+\;2\cos 6° \sin 12°}$ 
$\Large{1\over T} = {1\over 2\cos 6° \cos 12°} + \normalsize \tan 12°$
$\begin{align} \tan 36° - \tan12° 
&= {\sin36° \over \cos36°} - {\sin12° \over \cos12°} \cr
&= {\sin36° \cos12° - \cos36° \sin12° \over \cos12° \cos36°} \cr
&= {\sin24° \over \cos12° (\cos72° + 0.5)} \cr
&= {\sin24° \over \cos12°(\sin18° + \sin30°)} \cr
&= {\sin24° \over \cos12°(2 \sin24° \cos6°)} \cr
\tan36° &= {1\over 2\cos 6° \cos 12°} + \tan12°
\end{align}$
$$T = \cot 36° = \tan 54°$$
