A useful fact about triangles is $A = rs,$
where $r$ is the inradius and $s$ is the "semiperimeter"
$$s := \frac{a+b+c}{2}.$$
Why? Draw in the segments between the incenter and the vertices (these are the angle bisectors). This cuts the triangle into three smaller triangles. Each has height $r$ and the bases are $a,b,$ and $c$, respectively. So the areas of these triangles are $\frac{ra}{2}, \frac{rb}{2},$ and $\frac{rc}{2}$; the sum of these areas is $rs$.
Now that you have this; what's another way of writing the area of a right triangle? We'd have $A = \frac{1}{2}(2mn)(m^2-n^2) = mn(m^2-n^2) = mn(m-n)(m+n)$. For this triangle, $s = \frac{1}{2}(2mn + (m^2-n^2) + (m^2+n^2)) = mn + m^2 = m(n+m)$.
Equating our two expressinos for the area,
$mn(m-n)(m+n) = rm(n+m)$, and cancelling gives $r = n(m-n) = mn - n^2$.