<s>[INST] Show differential equation solution is arc of great circle I have to show that the solution  of a differential equation is an arc of a great circle. The differential equation is as follows $($in spherical coordinates$)$: 
$$\frac{\sin ^2\theta\phi '}{(1+\sin ^2\theta (\phi ')^2)^{\frac{1}{2}}}=C$$ where $C$ is an arbitrary constant and $\phi '$ denotes the derivative of $\phi$ with respect to $\theta$.
My reasoning: 
By setting $\phi (0)=0$, any arc of a great circle will have no change in $\phi$ with respect to $\theta$, so with this initial condition the answer follows by proving that $\phi '=0$. My issue is that upon working this round I end up with $$(\phi ')^2=\frac{C^2}{\sin ^4\theta -C^2\sin ^2\theta}$$ From this I can see no way forward.
Where do i go from here/ what should I do instead?
 [/INST] Let $u=\cot \theta$, then $1+u^2=\csc^2 \theta$ and $du=-\csc^2 \theta \, d\theta$.
\begin{align}
  \frac{d\phi}{d\theta} &= \frac{d\theta}{\sin \theta \sqrt{\sin^2 \theta-C^2}} \\
  &= \frac{C\csc^2 \theta}{\sqrt{1-C^2\csc^2 \theta}} \\
  d\phi &= -\frac{C\,du}{\sqrt{1-C^2(1+u^2)}} \\
  &= -\frac{C\,du}{\sqrt{(1-C^2)-C^2u^2}} \\
  &= -\frac{du}{\sqrt{\tan \alpha^2-u^2}} \tag{$C=\cos \alpha$} \\
  \phi &=\cos^{-1} \left( \frac{u}{\tan \alpha} \right)+\beta \\
  \cos (\phi-\beta) &= \frac{\cot \theta}{\tan \alpha} \\
  \cot \theta &= \tan \alpha \cos (\phi-\beta) \\
\end{align}
Rearrange,
$$(\sin \theta \cos \phi)(\sin \alpha \cos \beta)+
  (\sin \theta \sin \phi)(\sin \alpha \sin \beta)=
  (\cos \theta)(\cos \alpha)$$
which lies on the plane
$$x\sin \alpha \cos \beta+y\sin \alpha \sin \beta-z\cos \alpha=0$$
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