Generally, the radiative transfer equation is given by
$$
\frac{1}{c}\frac{\partial I_{\nu}(\vec{r},\vec{n},t)}{\partial t} +
\vec{n}\cdot\frac{I_{\nu}(\vec{r},\vec{n},t)}{\partial \vec{r}} =
\rho(\vec{r},t) \kappa_{\nu} (\vec{r},t)\left\{-I_{\nu}(\vec{r},\vec{n},t)+S_{\nu}(\vec{r},\vec{n},t)\right\}.
$$
To monochromatic intensity is defined by noting that $I_{\nu}(\vec{r},\vec{n},t)\cos\!\Theta \,d\nu\,df d\Omega\, dt$ at the position $\vec{r}$ gives the radiative energy
in the frequency interval $\nu,\nu+d\nu$ transported through the surface element $df$ into the solid
angle element $d\Omega$ enclosing the direction $\vec{n}$ during a time increment $dt$,
see the appended figure.
Your definition looks a bit strange, there is probably a time differential $dt$ missing in the denominator and the $2\phi$ should be the energy differential $dE$. However, $\cos\theta$ in your definition of the radiance correctly appears in the denominator. Looking at my figure, if $\Theta = 90^{\circ}$ the energy propagates along the area element $df$ instead of though it, so the radiance is indeed ill defined in this case.