Definig a angle between two vectors $a,c$ from a hilbert space $\mathcal{H}$ by $$\theta(a,c) = \arccos\frac{\mathfrak{Re}( \langle a,c\rangle)}{\|a\|\|c\|} $$ or by: $$\widetilde{\theta}(a,c) = \arccos\frac{ |\langle a,c\rangle|}{\|a\|\|c\|} $$
Then if $\{a,b \}$ is a orthonormal basis of $\mathcal{H}$ and hence $\langle a,b \rangle =0 \Rightarrow \theta(a,b)= \widetilde{\theta}(a,b) = \frac{\pi}{2}$
Is the assertion that if for a direction given by a unit vector $c$, and a angle $\theta(c,a)$ $($or possibly $\widetilde{\theta}(c,a))$, then does that mean that $\theta(c,b)=\theta(c,a) +\frac{\pi}{2}$ ? $\left(\text{equivalently } \widetilde{\theta}(c,b)=\widetilde{\theta}(c,a) +\frac{\pi}{2} \,?\right)$
This woud make sense if we could calculate $\theta(c,b)-\theta(c,a)=\theta(a,b) $ or $\widetilde{\theta}(c,b)-\widetilde{\theta}(c,a) =\widetilde{\theta}(a,b)$, but the arccos of either $|\cdot|$ or $\mathfrak{Re}(\cdot)$ isn't linear and it becomes crumbly trying to prove those relations, I suspect that although the Hilbert space is two dimensional, the fact that the scalar are complex, does not allow for a simple sum and subtraction of angles as it would be in a two dimensional Euclidean space.
This question comes about because I was reading a article in which one supposedly shows a concrete conterexample of the Kochen-Specken theorem for two dimensional Hilbert spaces and suddenly begins to "re-describe" the vectors in this two dimensional Hilbert space by real angles to a fixed $x$ diection, these angles then would be summed and subtracted, and since I have never seen anything like this I am very skeptical if that is allowed.