Is it possible to "re-normalize" a Dirac delta function? The delta function in spherical coordinates is given by:
$$\delta(\vec{r}_0-\vec{r})=\frac{1}{r^2}\delta(r_0-r)\delta(\cos\theta_0-\cos\theta)\delta(\phi_0-\phi),$$
(The ordering of the coordinates inside the $\delta$'s isn't important). If I have a particular location in mind, say $(r_0,\theta_0,\phi_0)=(r_0,0,0)$, is there a neat way to "re-normalize" the following:
$$\cot\theta_0\left(\frac{1}{r^2}\delta(r_0-r)\delta(\cos\theta_0-\cos\theta)\delta(\phi_0-\phi)\right),$$
since $\cot\theta_0$ is also singular at $\theta_0\in \pi \mathbb{Z}$.
To be clear, I'm trying to see if there is a way to cleverly absorb this cotangent function into the delta function so that it remains relatively unchanged. It's also possible this question is entirely unfounded, and I apologize if so. 

It seems OP wants to avoid having poles inside & outside the Dirac delta distribution. Using a hopefully obvious notation 
$$s~\equiv~ \sin\theta,\qquad c~\equiv~ \cos\theta,$$
one possibility is to replace
$$ \frac{c}{s} \delta(c-c_0)~=~ \frac{sc}{|f^{\prime}(c)|} \delta(c-c_0)~=~sc~\delta(f(c)-f(c_0)), $$
where $$f(c)~:=~c(1-\frac{c^2}{3}),\qquad f^{\prime}(c) ~=~1-c^2, \qquad c~\in~[-1,1].$$
Of course such rewritings are essentially just cosmetics. The singular nature of the construction does not disappear but manifests itself in other ways. 
