In a traingle ABC, C = ${90}^0$. Then $\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2}=$
 

$\begin{array}{rcl}Given \angle \mathrm{C}& =& {90}^0 \Rightarrow \mathrm{A}+\mathrm{B}={90}^0\\ Now \frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2}& =& \frac{4{\mathrm{R}}^2 {\sin }^2\mathrm{A}-4{\mathrm{R}}^2 {\sin }^2\mathrm{B}}{4{\mathrm{R}}^2 {\sin }^2\mathrm{A}+4{\mathrm{R}}^2 {\sin }^2\mathrm{B}}\\ & =& \frac{{\sin }^2\mathrm{A}-{\sin }^2\mathrm{B}}{{\sin }^2\mathrm{A}+{\sin }^2\mathrm{B}}\\ & =& \frac{\sin (\mathrm{A}+\mathrm{B})\sin (\mathrm{A}-\mathrm{B})}{{\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}}\\ & =& \sin (\mathrm{A}-\mathrm{B})\end{array}$