If triangle ABC is a non-isosceles triangle a | vedclass

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(A) Given that in $	riangle ABC$,$\angle C = 90^{\circ}$.Since $\angle A + \angle B + \angle C = 180^{\circ}$,we have $\angle A + \angle B = 90^{\cir

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(A) Given that in $	riangle ABC$,$\angle C = 90^{\circ}$.Since $\angle A + \angle B + \angle C = 180^{\circ}$,we have $\angle A + \angle B = 90^{\circ}$.Using the sine rule,$a = k \sin A$ and $b = k \sin B$.Substituting these into the expression:$\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = \frac{k^2 \sin^2 A + k^2 \sin^2 B}{k^2 \sin^2 A - k^2 \sin^2 B} \sin(A-B) = \frac{\sin^2 A + \sin^2 B}{\sin^2 A - \sin^2 B} \sin(A-B)$.Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$:$= \frac{\sin^2 A + \sin^2 B}{\sin(A+B) \sin(A-B)} \sin(A-B) = \frac{\sin^2 A + \sin^2 B}{\sin(A+B)}$.Since $A+B = 90^{\circ}$,$\sin(A+B) = 1$ and $B = 90^{\circ}-A$,so $\sin B = \cos A$.$= \frac{\sin^2 A + \cos^2 A}{1} = 1$.

If $\triangle ABC$ is a non-isosceles triangle and $\angle C = 90^{\circ}$,then $\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = $

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