$\angle C=90^{\circ}, \frac{a^2-b^2}{a^2+b^2}$ ...(i)
$\because \quad \cos c=\frac{a^2+b^2-c^2}{2 a b}=\cos 90^{\circ}$
$\Rightarrow \quad \frac{a^2+b^2-c^2}{2 a b}=0$
$a^2+b^2-c^2=0$
$\Rightarrow \quad c^2=a^2+b^2$
From Eq. (i), we get
$\frac{a^2-b^2}{a^2+b^2}=\frac{a^2-b^2}{c^2}$
$\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=R$ (constant)
$=\frac{(R \sin A)^2-(R \sin B)^2}{(R \sin C)^2}=\frac{\sin ^2 A-\sin ^2 B}{\sin ^2 90^{\circ}}$
$\left(\because \sin 90^{\circ}=1\right)$
$=(\sin A+\sin B)(\sin A-\sin B)$
$=2 \cdot \sin \frac{A+B}{2} \cdot \cos \frac{A-B}{2} \cdot 2 \cos \frac{A+B}{2}$ $\cdot \sin \frac{A-B}{2}$
$=2 \sin \frac{\pi}{4} \cdot \cos \frac{A-B}{2} \cdot 2 \cos \frac{\pi}{4} \cdot \sin \frac{A-B}{2}$
$\because A+B=\pi-C$ $=\pi-\frac{\pi}{2}=\frac{\pi}{2}$
$=2 \cdot 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \cdot \sin \frac{A-B}{2} \cdot \cos \frac{A-B}{2}$
$=2 \cdot \sin \left(\frac{A-B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)$
$\because \sin 2 A=2 \sin A \cdot \cos A$
$=\sin \left[2 \times \frac{(A-B)}{2}\right]$
$=\sin (A-B)$
Hence, $\quad \frac{a^2-b^2}{a^2+b^2}=\sin (A-B)$