Condition for 4 points on a hyperbola to be concyclic

Question

I found a post claiming that if $\alpha,\beta,\gamma,\theta$ are the eccentric angles of points $X,Y,Z,T$ we can parameterize those points on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ as $(a\sec(t),b\tan(t))$ which makes sense. Further in the post is proven that if the four points are concyclic we have $\tan(\frac{\alpha+\beta+\gamma+\theta}{2})=0$. My question is whether this works backwards. I checked on desmos and almost never points with eccentric angles $\alpha,\beta,\gamma, -\alpha-\beta-\gamma$ are concyclic. Can someone provide a reason for this.