 Solution

Because $k>0$ and the original smooth closed curve has periodic Frenet data, $\arctan (\tau /k)$ is a continuous real-valued periodic function. Therefore $\int k_{g}d\bar{s}=-\oint d\arctan (\tau /k)=0$. The simple closed normal curve divides the sphere into two disks by the Jordan theorem. Apply Gauss-Bonnet theorem to either disk, choosing its consistent boundary orientation. Its Gaussian curvature is one and its total boundary curvature remains zero. Thus each region has area $2\pi$, proving the two regions have equal area.