In triangle ABC,if angle C = 90^{circ},then f | vedclass

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(A) Given that $\angle C = 90^{\circ}$,we have $A + B = 90^{\circ}$,which implies $B = 90^{\circ} - A$. Substituting this,we get $\sin B = \sin(90^{\c

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(A) Given that $\angle C = 90^{\circ}$,we have $A + B = 90^{\circ}$,which implies $B = 90^{\circ} - A$. Substituting this,we get $\sin B = \sin(90^{\circ} - A) = \cos A$. Thus,$\sin^2 B = \cos^2 A$. The expression becomes $\frac{\sin^2 A + \cos^2 A}{\sin^2 A - \cos^2 A} \sin(A - (90^{\circ} - A))$. Since $\sin^2 A + \cos^2 A = 1$ and $\sin^2 A - \cos^2 A = -\cos(2A)$,the expression is $\frac{1}{-\cos(2A)} \sin(2A - 90^{\circ})$. Using $\sin(2A - 90^{\circ}) = -\cos(2A)$,the expression simplifies to $\frac{1}{-\cos(2A)} 	imes (-\cos(2A)) = 1$.

In $\triangle ABC$,if $\angle C = 90^{\circ}$,then $\frac{(\sin^2 A + \sin^2 B)}{(\sin^2 A - \sin^2 B)} \sin(A - B) = $

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