If ABC is a right-angled triangle with 90^{ci | vedclass

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(B) Using the sine rule,we have $a = 2R \sin A$ and $b = 2R \sin B$. Substituting these into the expression:$\frac{a^2+b^2}{a^2-b^2} \sin (A-B) = \fra

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$1$

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(B) Using the sine rule,we have $a = 2R \sin A$ and $b = 2R \sin B$. Substituting these into the expression:$\frac{a^2+b^2}{a^2-b^2} \sin (A-B) = \frac{(2R \sin A)^2 + (2R \sin B)^2}{(2R \sin A)^2 - (2R \sin B)^2} \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 $A+B+C = 180^{\circ}$ and $C = 90^{\circ}$,we have $A+B = 90^{\circ}$,so $\sin(A+B) = \sin 90^{\circ} = 1$ and $B = 90^{\circ} - A$,which implies $\sin B = \cos A$.$= \frac{\sin^2 A + \cos^2 A}{1} = 1$Thus,the correct option is $B$.

If $ABC$ is a right-angled triangle with $90^{\circ}$ at $C$ and $a > b$,then $\frac{a^2+b^2}{a^2-b^2} \sin (A-B) = $

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