(doCarmo, Exercise 2, Sec. 2-5) Let $ \textbf{x}(\phi,\theta) = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$ be a parametrisation of the unit sphere $S^2$. Let P be the plane $x=z\cot \alpha$, $0 < \alpha < \pi$, and $\beta$ be the acute angle which the curve $P \cap S^2$ makes with the semimeridian $\phi = \phi_0$. Compute $\cos \beta$.
I assumed the basis $\{{\textbf{x}_{\theta},\textbf{x}_{\phi}}\}$ and computed the components of the two curves in terms of these and used the standard formula for the inner product to get $$\cos \beta = \frac{1}{\sqrt{1+\cot^2 \alpha + \cos^2 \phi_0}}.$$ Would just like to confirm the answer with someone else - thanks in advance.
HINT:
Trying to check your result in alternate trig form
$$ \tan ^2 \beta = 1/ \tan ^2 \alpha + \cos ^2 \phi_0 ^2 $$
At equator meridian should cut great circle at complementary angles.
At max polar angle it should cut at angle $\beta = \pi/2. $
Best way is to check by using Claraut's Law, not so sure/comfortable about notation..
$$ \sin \theta \sin \beta = ..\cos (\alpha).. $$
EDIT1:
I derived to obtain a result which in your nomenclature is:
$$ \tan \beta = \frac{ \cot \alpha \sin \theta - \cos \theta \sin \phi }{ \cos \phi} $$
It should be checked and perhaps be further simplified, using your given conditions by division
$$ \cot \alpha = \tan \theta \cos \phi $$