You could use the usual substitution. I use ArcTan[y,x] rather than ArcTan[y/x] because it accounts for the quadrant.

* Below edited to correct error *

In[1]:= fromPolarRule = {r -> Sqrt[x^2 + y^2], \[Theta] -> 
    ArcTan[x,y]};

In[2]:= eq = r Cos[\[Theta]] == r Sin[\[Theta]];

In[3]:= eqxy = eq /. fromPolarRule

Out[3]= x == y

Perhaps CoordinateTransform?

CoordinateTransform["Polar" -> "Cartesian", {r, \[Theta]}]
(* => {r Cos[\[Theta]], r Sin[\[Theta]]} *)

TransformedField[ "Polar" -> "Cartesian", [(r^2)* Cos[[Theta]]], {r, [Theta]} -> {x, y}]

Syntax::sntxf: "TransformedField[Polar->Cartesian," cannot be followed by "[(r^2)*Cos[[Theta]]],{r,[Theta]}->{x,y}]". I am taking a calculus course on "Great Courses". On the Polar part of the course he shows how to convert an expression written in Polar Coordinate to Cartesian Coordinates. I still can't do this in Mathematica. I took the suggestion to try TransformedField. As you can see, I can't execute the example shown on the Help page. Do I have to load some kind of package before executing what I am trying to do. It doesn't look like it. A few years age I used Mathematica a lot. But now I am rusty. I my Trig book the show an example: r=@/(1-Cos(theta)) Converts to Y^2=4x+4. I have done polarplots ,successfully, in Mathematica on everything I am I am trying yo convert.

You help is appreciated.

Can I supply more info?

Well, after 20+ year using Mathematica I still learn something new every day. ;-)

The only thing wrong with your use of Transform field above is too many brackets, although code not in a code block sometimes gets mangled in a posting.

Below is a use of the built-in function you found, as well as an example of making your own, which is one of Mathematica's real strengths.

* and note that in my earlier post above I incorrectly used ArcTan[y,x] rather than ArcTan[x,y] *

In[1]:= (* with the built-in function *)
TransformedField[
 "Polar" -> "Cartesian", (r^2) Cos[theta], {r, theta} -> {x, y}]

Out[1]= x Sqrt[x^2 + y^2]

In[2]:= (* doing it with a Rule and Replace works also *)
fromPolarRule = {r -> Sqrt[x^2 + y^2], theta -> ArcTan[x, y]};

In[3]:= (* and we can use it to define a function *)
myTransformedField[expression_] := expression /. fromPolarRule

In[4]:= myTransformedField[(r^2) Cos[theta]]

Out[4]= x Sqrt[x^2 + y^2]

Thanks, Eric -- I edited it to correct the error.