After some calculations I have
$$\arctan\biggl(\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}\biggr)-\arcsin\biggl(\frac{3}{2\sqrt{4+\sqrt{3}}}\biggr).$$
Mathematica simplyfies this to $\pi/12$ and I wonder how it is done.
Edit: The results comes from calculating $x$ in the figure below. There might be simple ways to do it than mine.
I add my solution:
Cosinus Theorem:
$$
|AB|^2
=1^2+(\sqrt{3}\,)^2-2\cdot1\cdot\sqrt{3}\cdot\cos(120°)
=1+3-2\sqrt{3}(-\tfrac{1}{2})
=4+\sqrt{3}.
$$
Pythagoras:
$$
|AC|^2+|AB|^2=|BC|^2
\quad\Leftrightarrow\quad
|AC|^2
=|BC|^2-|AB|^2
=4+\sqrt{3}-(\sqrt{2}\,)^2
=4+\sqrt{3}-2
=2+\sqrt{3}
$$
i.e.
$$\textstyle|AC|=\sqrt{2+\sqrt{3}}.$$
Sinus Theorem
$$
\frac{\sin(y)}{\sqrt{3}}=\frac{\sin(120°)}{\sqrt{4+\sqrt{3}}}
\quad\Leftrightarrow\quad
\frac{\sin(y)}{\sqrt{3}}=\frac{\frac{\sqrt{3}}{2}}{\sqrt{4+\sqrt{3}}}
\quad\Leftrightarrow\quad
\sin(y)
=\frac{3}{2\sqrt{4+\sqrt{3}}}
$$
i.e.
$$
y=\arcsin\biggl(\frac{3}{2\sqrt{4+\sqrt{3}}}\biggr)
$$
Further,
$$
\tan(x+y)
=\frac{|AC|}{|AB|}
=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}
$$
i.e.
$$
x+y=\arctan\biggl(\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}\biggr).
$$
and
$$
x=\arctan\biggl(\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}\biggr)-\arcsin\biggl(\frac{3}{2\sqrt{4+\sqrt{3}}}\biggr)
$$
Best Answer
Try to take some trig function of all expression. For example, $\tan$: $$\tan\left(\underbrace{\arctan \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2}}}_{=:\alpha} - \underbrace{\arcsin\frac{3}{2\sqrt{4+\sqrt{3}}}}_{=:\beta}\right)=$$ $$=\tan(\alpha-\beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\cdot\tan\beta}$$ There is no problem is computing $\tan\alpha$ because it is $\tan(\arctan(\cdot))$. We need to find $\tan(\arcsin(\cdot))$: $$\tan(\arcsin x) = \{x = \sin\gamma\}=\tan\gamma = \frac{\sin\gamma}{\cos\gamma} = \frac{x}{\pm\sqrt{1-x^2}}$$ Plus or minus choice depends on $\gamma$. After that you will know $\tan$ of the angle which should be equal to $$\tan\frac{\pi}{12} = \frac{\sin\frac{\pi}{12}}{\cos\frac{\pi}{12}} = \frac{2\sin\frac{\pi}{12}\cos\frac{\pi}{12}}{2\cos^2\frac{\pi}{12}}= \frac{\sin\frac{\pi}{6}}{1+\cos\frac{\pi}{6}} = \frac{1}{2 + \sqrt{3}}=2 - \sqrt{3}$$