In a Delta ABC,if C = 90^{circ},then frac{a^{ | vedclass

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(B) Given $\angle C = 90^{\circ}$,in $\Delta ABC$,we have $a = c \sin A$ and $b = c \sin B$.Substituting these into the expression:$\frac{a^{2} + b^{2

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(B) Given $\angle C = 90^{\circ}$,in $\Delta ABC$,we have $a = c \sin A$ and $b = c \sin B$.Substituting these into the expression:$\frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin(A - B) = \frac{c^{2} \sin^{2} A + c^{2} \sin^{2} B}{c^{2} \sin^{2} A - c^{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 $\sin^{2} A - \sin^{2} B = \sin(A + B) \sin(A - B)$:$= \frac{\sin^{2} A + \sin^{2} B}{\sin(A + B) \sin(A - B)} \sin(A - B)$Since $A + B = 90^{\circ}$,$\sin(A + B) = \sin 90^{\circ} = 1$ and $\sin A = \cos B$:$= \frac{\cos^{2} B + \sin^{2} B}{1} = 1$.

In a $\Delta ABC$,if $C = 90^{\circ}$,then $\frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin(A - B)$ is equal to:

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