Hello Tom, Perhaps I am still not getting it. I am guessing that you are using parenthesis above for two different things (grouping and functions; functions use a square bracket notation) But, doesn't this little experiment suggest that you will not get a unique solution?

Combining both of your equations:

eq[g_, t_, q_] := (1 + t) g[t, q] == (1 + 1/q) g[1/q, 1/t]

Example arbitrary function of two variables:

ftest1[x_, y_] := (y^2/x)

ftest2[x_, y_] := (x^2/y)

eq[ftest1, t, q]

eq[ftest2, t, q]

eq[ArcTan, t , q]

With[

{equation1 = eq[ftest1, t, q], equation2 = eq[ArcTan, t , q],

equation3 = eq[ftest2, t, q]},

ContourPlot[{equation1, equation2, equation3}, {t, -10, 10}, {q, -10,

10}, ContourStyle -> {{Red, Thickness[0.02], Dashed}, {Blue,

Thickness[0.01]}, {Green}}]

]

Seems to show that there common values of t and q are satisfied simultaneously by three different functions.