= Solution
Take the unhalved geodesic <Lagrangian>
$$
\mathcal L=-c^2\alpha\dot t^2+\alpha^{-1}\dot r^2+r^2\dot\theta^2+r^2\sin^2\theta\dot\phi^2.
$$
The cyclic coordinates $t,\phi$ give $d(2c^2\alpha\dot t)/d\lambda=0$ and $d(2r^2\sin^2\theta\dot\phi)/d\lambda=0$. The other two <Euler-Lagrange equations> are
$$
\frac d{d\lambda}(2r^2\dot\theta)-2r^2\sin\theta\cos\theta\dot\phi^2=0,
$$
$$
\frac d{d\lambda}\left(\frac{2\dot r}{\alpha}\right)+c^2\alpha'\dot t^2+\frac{\alpha'}{\alpha^2}\dot r^2-2r\dot\theta^2-2r\sin^2\theta\dot\phi^2=0.
$$
These reproduce the supplied geodesic equations. Expanding the derivatives and comparing with $\ddot x^\mu+\Gamma^\mu{}_{\nu\rho}\dot x^\nu\dot x^\rho=0$, in coordinates $(t,r,\theta,\phi)$, gives all nonzero <Christoffel symbols>:
$$
\boxed{\begin{aligned}
\Gamma^t{}_{tr}=\Gamma^t{}_{rt}&=\frac{\alpha'}{2\alpha},&
\Gamma^r{}_{tt}&=\frac{c^2\alpha\alpha'}2,&
\Gamma^r{}_{rr}&=-\frac{\alpha'}{2\alpha},\\
\Gamma^r{}_{\theta\theta}&=-r\alpha,&
\Gamma^r{}_{\phi\phi}&=-r\alpha\sin^2\theta,&
\Gamma^\theta{}_{r\theta}=\Gamma^\theta{}_{\theta r}&=\frac1r,\\
\Gamma^\phi{}_{r\phi}=\Gamma^\phi{}_{\phi r}&=\frac1r,&
\Gamma^\theta{}_{\phi\phi}&=-\sin\theta\cos\theta,&
\Gamma^\phi{}_{\theta\phi}=\Gamma^\phi{}_{\phi\theta}&=\cot\theta.
\end{aligned}}
$$
Here $\alpha'=-2r/r_S^2$. This is the <static patch of de Sitter spacetime>, with $0\leq r<r_S$ for the stationary observers used below. Apparent divergences at the polar axis, origin or horizon belong to the chosen coordinate chart; they are not themselves evidence of a curvature singularity.
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