How do you find two functions $f$ and $g$ such that $f(x) \cdot g(x)=f(x)-g(x)$?

Question

This was inspired by this question ( Logarithms with trigonometric inequality )

I already know the answer ( $f(x)=\tan^2 x$ and $g(x)=\sin^2 x$). However I am interested in how to find this answer. Is there any way other that guessing? If not, what would probably be the least extra information you would need?

P.S. This looks like a potentially VERY useful identity. Is it well known and used?

Answer 1

This is nothing special. Pick any function $f(x)$ and put $$g(x)=\frac{f(x)}{1+f(x)}$$ with the usual caveats about domain.

Answer 2

$$ f (x)\cdot g (x)=f (x)-g (x)\\ \implies \dfrac {f (x)}{g (x)}-1=f (x)\\ \implies f (x)=\dfrac {1}{\dfrac {1}{g (x)}-1} $$ Then, let $ f (x)=\tan^2 (x)=\dfrac {\sin^2 (x)}{\cos^2 (x)}$. Substituting gives $$f (x)=\dfrac {1}{\dfrac {1}{g (x)}-1}\\ \implies f (x)=\sin^2 (x)$$ Of course, you consider the appropriate domains.