#### QUESTION

This seems like a simple thing to do, but I couldn't find anything relevant from Mathematica documentation.

So suppose I have an expression:

```
a*b/(a + a*Cos[a/b])
```

And I have defined:

```
k=a/b
```

Now I want to simplify the expression above so that the simplify would use my definition of k in place of a/b in as many places as possible so that the final expression would look something like:

```
a/(k+k*Cos[k])
```

This was just a simple example I made up to demonstrate what I'd like to do, but I have encountered a similar situations many times every now and then.

#### ANSWER

Daniel Lichtblau and Andrzej Koslowski posted a solution in mathgroup, which I adjusted marginally. (I like to use german identifiers, because they will never clash with Mma builtins). That's the code:

```
SetAttributes[termErsetzung,Listable];
termErsetzung[expr_, rep_, vars_] :=
Module[{num = Numerator[expr], den = Denominator[expr],
hed = Head[expr], base, expon},
If[PolynomialQ[num, vars] && PolynomialQ[den, vars] && ! NumberQ[den],
termErsetzung[num, rep, vars]/termErsetzung[den, rep, vars], (*else*)
If[hed === Power && Length[expr] === 2,
base = termErsetzung[expr[[1]], rep, vars];
expon = termErsetzung[expr[[2]], rep, vars];
PolynomialReduce[base^expon, rep, vars][[2]], (*else*)
If[Head[Evaluate[hed]] === Symbol &&
MemberQ[Attributes[Evaluate[hed]], NumericFunction],
Map[termErsetzung[#, rep, vars] &, expr], (*else*)
PolynomialReduce[expr, rep, vars][[2]] ]]]
];
TermErsetzung[rep_Equal,vars_][expr_]:=
termErsetzung[expr,Evaluate[Subtract@@rep],vars]//Union;
```

Usage is like this:

```
a*b/(a + a*Cos[a/b]) // TermErsetzung[k b == a, b]
```

a/(k (1 + Cos[k]))

The first parameter is the "replacement equation", the second the variable (or list of variables) to be eliminated:

```
a*b/(a + a*Cos[a/b]) // TermErsetzung[k b == a, {a, b}]
```

Tweet{b/(1 + Cos[k]), a/(k (1 + Cos[k]))}