Because of convenient trigonometric identities, we can find the exact value of things like $\tan70^{\circ}$ because $70^{\circ}=2\cdot35^{\circ}$ (tangent double-angle identity), $35^{\circ}=36^{\circ}-1^{\circ}$ (tangent angle difference identity), $3^{\circ}=3\cdot1^{\circ}$ (tangent triple-angle identity) and $3^{\circ}=18^{\circ}-15^{\circ}$ (tangent angle difference identity), $18^{\circ}=\frac{1}{2}\cdot36^{\circ}$ (tangent half-angle identity), and $15^{\circ}=\frac{1}{2}\cdot30^{\circ}$, leaving $\tan70^{\circ}$ in terms of the more well-known $\tan30^{\circ}$ and $\tan36^{\circ}$. I am fairly sure that we can find exact values by following similar procedures for all $\tan\theta$ ($\theta$ in degrees) as long as $\theta\in\mathbb{Q}$.
So, can we do the same for finding the exact values of inverse trigonometric functions? Specifically, I am looking for a method of finding the exact value of $\tan^{-1}\frac{1}{2}$ in terms of things like integers, radicals, and complex numbers (which are required in the case of $\tan70^{\circ}$, for example), but not other trigonometric functions or infinite series. I am not asking for the exact value itself as I am sure no one would like to burden themselves finding it, but rather just the method so that I may.
Just an illustration (I cannot post this as comment):
$$t_3=\tan3^\circ=\frac{-2 \sqrt{2 \left(\sqrt{5}+3\right)}+\sqrt{6 \sqrt{5}+15}+1}{\sqrt{3}-\sqrt{2 \sqrt{5}+5}-2 \sqrt{4 \sqrt{5}-\sqrt{2 \left(\sqrt{5}+3\right)}-\sqrt{66 \sqrt{5}+150}+12}}$$
$$t_1=\tan1^\circ=-\frac{1}{2} \left(1+i \sqrt{3}\right) \sqrt[3]{t_3^3+\sqrt{-\left(t_3^2+1\right)^2}+t_3}+\frac{i \left(\sqrt{3}+i\right) \left(t_3^2+1\right)}{2 \sqrt[3]{t_3^3+\sqrt{-\left(t_3^2+1\right)^2}+t_3}}+t_3 $$
So, yes, you can calculate $\tan N^\circ$ for any natural number N exactly.