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}}$?

$\begin{align} T &= \large \left(\frac{2\cos 6° \cos 12°}{1\;+\;2\cos 6° \sin 12°}\right) \left({2\sin 6° \over 2\sin 6°}\right)\cr &= \large \frac{\sin 24°}{2\sin 6°+\;2\sin^2 12°}\cr\cr {1\over T}&= \large \frac{2\sin(30°-24°)+\;(1-\cos 24°)}{\sin 24°}\cr &=\large {(\cos 24° - \sqrt3\sin24°) + (1-\cos 24°) \over \sin 24°}\cr &=\large{1\over\sin(60°-36°)} - \sqrt3 \cr &=\large{4\over 2\sqrt3 \cos36° - 2\sin 36°}-\sqrt3 \cr \end{align}$

From wikipedia on Golden triangle, $\quad2\cos(36°)=\phi,\quad 2\sin(36°)=\sqrt{4-\phi^2}$

$\begin{align} \large{1\over T}&=\left({4\over \sqrt3\phi-\sqrt{4-\phi^2}}\right) \left({\sqrt3\phi+\sqrt{4-\phi^2} \over \sqrt3\phi+\sqrt{4-\phi^2}} \right) -\sqrt3 \cr &= {4(\sqrt3\phi + \sqrt{4-\phi^2}) \over 3\phi^2-(4-\phi^2)} -\sqrt3 \cr &= {\sqrt3\phi + \sqrt{4-\phi^2} \over \phi^2-1}-\sqrt3 \cr &= {\sqrt3\phi + \sqrt{4-\phi^2} \over \phi} -\sqrt3 \cr &= {\sqrt{4-\phi^2} \over \phi} = \tan(36°) \end{align}$

$$T = \tan(90°-36°) = \tan(54°)$$

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!