Given that, in $\triangle A B C$,
$\begin{aligned} \angle C & =90^{\circ} \\ \angle A+\angle B+\angle C & =180^{\circ} \\ \therefore \quad \angle A+\angle B & =90^{\circ}\end{aligned}$
Then, $\frac{a^2+b^2}{a^2-b^2} \sin (A-B)$
Using sine rule,we get
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$
$\therefore \quad \frac{(k \sin A)^2+(k \sin B)^2}{(k \sin A)^2-(k \sin B)^2} \sin (A-B)$
$\begin{aligned}= & \frac{\sin ^2 A+\sin ^2 B}{\sin ^2 A-\sin ^2 B} \sin (A-B) \\ = & \frac{\frac{1-\cos 2 A}{2}+\frac{1-\cos 2 B}{2}}{\frac{1-\cos 2 A}{2}-\frac{1-\cos 2 B}{2}} \sin (A-B) \\ = & \frac{1-\cos 2 A+1-\cos 2 B}{1-\cos 2 A-1+\cos 2 B} \sin (A-B) \\ = & \frac{2-\cos 2 A-\cos 2 B}{\cos 2 B-\cos 2 A} \sin (A-B) \\ = & \frac{2-(\cos 2 A+\cos 2 B)}{\cos 2 B-\cos 2 A} \sin (A-B) \\ = & \frac{2-\left(2 \cos \frac{(2 A+2 B)}{2} \cos \frac{(2 A-2 B)}{2}\right)}{2 \sin \frac{(2 B+2 A)}{2} \sin \frac{(2 A-2 B)}{2}} \cdot \sin (A-B) \\ = & \frac{2-(2 \cos (A+B) \cos (A-B))}{2 \sin (A+B) \sin (A-B)} \cdot \sin (A-B) \\ = & \frac{2-(2 \cos (A+B) \cos (A-B))}{2 \sin (A+B)} \\ = & \frac{2-\left(2 \cos 90^{\circ} \cos (A-B)\right)}{2 \sin 90^{\circ}}=\frac{2-0}{2}=\frac{2}{2}=1\end{aligned}$