Divergence of spherically symetric vector field not the same as in polar coordinates?

Question

I'm a bit confused here. The divergence of a vector in polar spherical coordinates is $$ \nabla\cdot\vec{A}=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2A_r)+\frac{1}{r\sin{\theta}}\frac{\partial}{\partial\theta}(A_{\theta}\sin{\theta})+\frac{1}{r\sin{\theta}}\frac{\partial A_{\phi}}{\partial\phi}, $$ while the divergence in polar coordinates is $$ \nabla\cdot\vec{A}=\frac{1}{r}\frac{\partial}{\partial r}(rA_r)+\frac{1}{r}\frac{\partial A_{\theta}}{\partial\theta}. $$ What confuses me, is why the latter is not just the former minus the $\phi$ component. In particular, I am interested in a problem that is initially in spherical coordinates. $\vec{A}$ is spherically symetric and thus in any case the last term of the first expression disappears. But I am interested in what is happening in meridional planes. So $\vec{A}$ is separated in a component in $r$ and $\theta$ and a component in $\phi$ alone. So the meridional component exists for all $\phi$, but they just don't vary with it. Now I always assumed that in that case, the meridional component can always be treated as if we were in polar coordinates, since I just ignore the $\phi$ component. So my question is this, I suppose I have to use the former expression but without the last term? And why doesn't the problem not simply reduce to one in polar coordinates?