How to make Simplify[] work with substitutions?

Question

There is a matrix $M$, and as far as it has quite a complicated form, there are several algebraic equations defining some of its parameters. I try to find the inverse of $M$ and simplify it. Unfortunately, $Mathematica$ gives the answer in the form, differing from the one I want: precisely, it substitutes parameters in the inverse matrix instead of giving the answer in terms of parameters (where it is possible).

Concretely, I have $$\rho = \sqrt{r^2 + a^2 Cos[\Theta]^2}$$ $$\Delta = r^2 - 2 m r + a^2 + e^2$$ and want the answer to be given in terms of $\Delta$ and $\rho$, where it is possible.

I tried to use $\mathsf{Eliminate[]}$ instead of $\mathsf{FullSimplify[]}$, but did not succeed. Moreover, there is a problem with $\mathsf{Eliminate[]}$: there may be no possibility to fully eliminate the parameters.

Is there any way to make $Mathematica$ work as I want?

EDIT:

I have following problem:

\[Rho] = Sqrt[r^2 + a^2 Cos[\[Theta]]^2];

\[CapitalDelta] = r^2 - 2 m r + a^2 + e^2;

Metric = {{\[Rho]^2, 0, 0, 0}, {0, 0, - a Sin[\[Theta]]^2, 
1}, {0, - 
  a Sin[\[Theta]]^2, \[Rho]^-2 ((r^2 + 
     a^2)^2 - \[CapitalDelta] a^2  Sin[\[Theta]]^2) \
Sin[\[Theta]]^2, - a \[Rho]^-2 (2 m r - e^2) Sin[\[Theta]]^2}, {0, 
1, - a \[Rho]^-2 (2 m r - 
   e^2) Sin[\[Theta]]^2, -(1 - \[Rho]^-2 (2 m r - e^2))}};

InvMetric = Inverse[Metric] // Simplify // MatrixForm

And what I get is:

,

although I want to get $\rho^{-2}$ instead of (0; 0) entry, for example.

Answer 1

The following works for your example:

Simplify[
    InvMetric,
    Cos[2 θ]==2Cos[θ]^2-1 && ρ^2==r^2+a^2 Cos[θ]^2 && Δ==r^2-2 m r+a^2+e^2
] //TeXForm

$\left( \begin{array}{cccc} \frac{1}{\rho ^2} & 0 & 0 & 0 \\ 0 & \frac{\Delta }{\rho ^2} & \frac{a}{\rho ^2} & \frac{a^2+r^2}{\rho ^2} \\ 0 & \frac{a}{\rho ^2} & \frac{\csc ^2(\theta )}{\rho ^2} & \frac{a}{\rho ^2} \\ 0 & \frac{a^2+r^2}{\rho ^2} & \frac{a}{\rho ^2} & \frac{a^2 \sin ^2(\theta )}{\rho ^2} \\ \end{array} \right)$