Q:

Prove that the loxodrome crosses all meridians at a constant angle

How to prove that the loxodrome (the rhumb line) crosses all meridians at a constant angle?
$$\tan\left(\frac{\pi}{4} + \frac{\psi}{2}\right) = e^{k\phi}, \quad k = \text{constant}$$
where  $\psi$ is the latitude of a point on a sphere, and $\phi$ is the longitude.

A:

The stereographic projection $\sigma:\>\dot S^2\to{\mathbb R}^2$ maps the point $P\in \dot S^2$with geographical cordinates $[\psi,\phi]$ to the point $\sigma(P)\in{\mathbb R}^2$ with polar coordinates $[r,\phi]$, whereby $r$ is given by
$$r={\cos\psi\over 1-\sin\psi}\ .$$
This is easily verified in a figure  showing a half meridian circle of $S^2$. Writing $\tan{\psi\over2}=:t$ we therefore have
$$r={{1-t^2\over 1+t^2}\over 1-{2t\over1+t^2}}={1+t\over1-t}=\tan\left({\pi\over4}+{\psi\over2}\right)\qquad\bigl(=e^{k\phi}\bigr)\ .$$
This shows that the stereographic image $\sigma(\ell)$ of your loxodrome is a logarithmic spiral $r=e^{k\phi}$ in the plane. Such a spiral intersects all radii originating at $(0,0)$ under the same angle $\alpha$ (which depends on $k$). Now these radii are just the stereographic images of the meridians on $S^2$. Since $\sigma$ is known to be conformal it follows that $\ell$ intersects all meridians on $S^2$ under the same angle $\alpha$.

