Solving this problem
Two identical circles passing through each other's centres. Three parallel lines and two diagonal lines drawn as below. What is the value of the marked angle?
I found that the request angle is $ \arccos [{\frac {1+3 \sqrt{3}}{2 \sqrt{10}}}] $, but this can be even written as $ \arctan [\frac {1} {2}] - \frac {\pi} {12} $ (you can see they are numerically equivalent).
How that equivalence can be proved with trigonometry?
On
Although you wanted a trigonometric/algebraic approach, for the sake of completeness, we will show how to compute the desired angle using a (mostly) geometric approach. We redraw and label the diagram as follows:
The desired angle is $\alpha = \angle F'FE$.
Claim 1. $\beta = \angle CFF' = \frac{\pi}{12}$.
Proof. Note that because circles $A$ and $B$ share a common radius $\overline{AB}$ and their centers pass through each other, that $F'A = AB = F'B$; hence $\triangle AF'B$ is equilateral, so $\angle F'AB = \frac{\pi}{3}$. Since $\angle CAB$ is right, then $$\angle CAF' = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}.$$ Therefore, by the inscribed angle theorem, $$\beta = \frac{\angle CAF'}{2} = \frac{\pi}{12},$$ since $\angle CAF'$ is a central angle subtending arc $CF'$, and $\angle CFF'$ is an inscribed angle subtending the same arc.
Claim 2. $\alpha + \beta = \gamma$.
Proof. This is simply another consequence of the inscribed angle theorem. Both inscribed $\angle CC'E$ and $\angle CFE$ subtend the same arc $CE$, so they are congruent.
Claim 3. $\gamma = \arctan \frac{1}{2}$.
Proof. We note $AO = \frac{1}{2} AB$ since $O$ is the midpoint of $\overline{AB}$. And since $\angle OAC'$ is right, we immediately have $$\tan \gamma = \frac{AO}{AC'} = \frac{\frac{1}{2} AB}{AB} = \frac{1}{2},$$ and the result follows.
Claim 4. $\alpha = \arctan \frac{1}{2} - \frac{\pi}{12}$.
Proof. From the above claims, $$\alpha = \gamma - \beta = \arctan \frac{1}{2} - \frac{\pi}{12}.$$
On
We will prove that $\;\color{brown}{\arccos\left(\!{\dfrac{1+3\sqrt3}{2\sqrt{10}}}\right)=\arctan\dfrac12-\dfrac\pi{12}}\,.$
Since $\;\cos\left(\dfrac\pi{12}\right)=\dfrac{\sqrt6+\sqrt2}4\;,\;$ $\;\sin\left(\dfrac\pi{12}\right)=\dfrac{\sqrt6-\sqrt2}4\;,$
$\cos\left(\arctan\dfrac12\right)=\dfrac1{\sqrt{1+\left(\frac12\right)^2}}=\dfrac2{\sqrt5}\;,\;\;\sin\left(\arctan\dfrac12\right)=\dfrac1{\sqrt5}\;,$
it follows that
$\begin{align}\arccos\left(\!{\dfrac{1+3\sqrt3}{2\sqrt{10}}}\right)&=\arccos\left[{\dfrac{\sqrt2\left(1+3\sqrt3\right)}{\sqrt2\cdot2\sqrt{10}}}\right]=\arccos\left(\dfrac{\sqrt2+3\sqrt6}{4\sqrt5}\right)=\\[3pt]&=\arccos\left[\dfrac{2\left(\sqrt6+\sqrt2\right)+\sqrt6-\sqrt2}{4\sqrt5}\right]=\\[3pt]&=\arccos\left(\dfrac2{\sqrt5}\!\cdot\!\dfrac{\sqrt6+\sqrt2}4+\dfrac1{\sqrt5}\!\cdot\!\dfrac{\sqrt6-\sqrt2}4\right)=\\[3pt]&=\arccos\left[\cos\left(\!\arctan\dfrac12\!\right)\cos\dfrac\pi{12}+\sin\left(\!\arctan\dfrac12\!\right)\sin\dfrac\pi{12}\right]=\\[3pt]&=\arccos\left[\cos\left(\arctan\dfrac12-\dfrac\pi{12}\right)\right]\underset{\overbrace{\text{ because }\arctan\!\frac12-\frac\pi{12}\,\in\,[0,\pi]\;\;}}{=}\\[0pt]&=\arctan\dfrac12-\dfrac\pi{12}\;.\end{align}$
Observe by the pythagorean we can express $\arccos{\left(\dfrac{1 + 3\sqrt{3}}{2\sqrt{10}}\right)}$ in terms of $\arctan$ function.
$$a^2 + b^2 = c^2 \iff b = \sqrt{c^2 - a^2} = \sqrt{(2\sqrt{10})^2 - (1 + 3\sqrt{3})^2} = \sqrt{12 - 6\sqrt{3}} = 3 - \sqrt{3} \\\therefore \arccos{\left(\dfrac{1 + 3\sqrt{3}}{2\sqrt{10}}\right)} = \arctan{\left(\dfrac{3 - \sqrt{3}}{1 + 3\sqrt{3}}\right)} = \arctan{\left(\dfrac{5\sqrt{3} - 6}{13}\right)}$$
Using the identity $\arctan{x} - \arctan{y} = \arctan{\dfrac{x - y}{1 + xy}}$, $xy < 1$
$$\arctan{\frac{1}{2}} - \arctan{\left(\dfrac{5\sqrt{3} - 6}{13}\right)} \\= \arctan{\left(\dfrac{\dfrac{1}{2} - \dfrac{5\sqrt{3} - 6}{13}}{1 + \dfrac{5\sqrt{3} - 6}{26}}\right)}$$
$\dfrac{\dfrac{1}{2} - \dfrac{5\sqrt{3} - 6}{13}}{1 + \dfrac{5\sqrt{3} - 6}{26}} = \dfrac{\dfrac{25 - 10\sqrt{3}}{26}}{\dfrac{5\sqrt{3} + 20}{26}} = \dfrac{25 - 10\sqrt{3}}{5\sqrt{3} + 20} = 2 - \sqrt{3}$
So, $$\arctan{(2 - \sqrt{3})} = \frac{\pi}{12}$$