We have

$\frac{a^2+b^2}{a^2-b^2}\sin (A-B)=1 \Rightarrow \sin (A-B)=\frac{a^2-b^2}{a^2+b^2} ...\left(1\right)$

Given, $0<B<45°$ and $\angle C=90°$

So, $0<A-B<90°$ $\Rightarrow 0<\sin \left(A-B\right)<1 \Rightarrow 0<\frac{a^2-b^2}{a^2+b^2}<1$  (from $\left(1\right)$)

$a^2-b^2>0 \Rightarrow a>b ...\left(2\right)$

Now, as, $C=90° \Rightarrow ∆ABC$ is a right-angled triangle whose hypotenuse is $c$.

Hence, $c>a>b$     (from $\left(2\right)$)