Question

In triangle ABC, if $C={90}^0$ then the value of $\frac{a^2+b^2}{a^2-b^2}sin(A-B)$ equals:

Answer 1

In $△ABC,C={90}^{\circ }⟹A+B={90}^{\circ }$

And also $a^2+b^2=c^2$

$\frac{a^2+b^2}{a^2-b^2}\sin (A-B)=\frac{c^2}{4R^2{\sin }^{2}A-4R^2{\sin }^{2}B}\sin (A-B)$

$=\frac{c^2}{4R^2({\sin }^{2}A-{\sin }^{2}B)}\sin (A-B)$

$=\frac{4R^2{\sin }^{2}C}{4R^2\sin (A+B)\sin (A-B)}\sin (A-B)$

$=\frac{{\sin }^2{90}^{\circ }}{\sin {90}^{\circ }}$

$=1$