The way I understood the question, the Cartesian parameters are fixed. I'm just going to call them $x_0$, $y_0$ and $z_0$. But since in spherical coordinates we also have $x=r\cos\phi \sin\theta$, $y=r\sin\phi\sin\theta$, $z=r\cos\theta$, we can rewrite the inequality as
$$0 \le x_0 x + y_0 y + z_0 z $$
(the radial coordinate $r$ cancels). This leads me to the following implementation using MeshFunctions
:
With[{x0 = -1/2, y0 = 0, z0 = 1/2},
SphericalPlot3D[1, θ, ϕ, Mesh -> {{0}}, PlotPoints -> 70,
MeshShading -> {None, Orange},
MeshStyle -> None,
MeshFunctions -> {Function[{x, y, z, θ, ϕ, r}, x x0 + y y0 + z z0]}]]

I used SphericalPlot3D
because it's fast. The mesh only has the contour 0
because that's the cutoff for the inequality. The formulation of the problem now uses only Cartesian coordinates, so I need just the corresponding first three arguments of the mesh function.
However, in multiplying the original inequality by trig functions to bring it into the simpler form above, some sign changes are lost. Thanks to theDude for pointing that out. The original inequality can easily be plotted in the same way, though:
With[{x0 = -1/2, y0 = 0, z0 = 1/2},
SphericalPlot3D[1, θ, ϕ, Mesh -> {{0}}, PlotPoints -> 70,
MeshShading -> {None, Orange}, MeshStyle -> None,
MeshFunctions -> {Function[{x, y, z, θ, ϕ, r},
Tan[θ] + z0 /(x0 Cos[ϕ] + y0 Sin[ϕ])]}]]

The plot agrees with the result of thedude's answer, but this method is faster.
RegionFunction -> Function[{\[Phi], \[Theta]}, Tan[\[Theta]] >= -z/(x Cos[\[Phi]] + y Sin[\[Phi]])]
$\endgroup$RegionFunction
will not work because it includes variables $x$, $y$, and $z$ along with $\theta$ and $\phi$. $\endgroup$