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

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(B) Given $\angle C = 90^{\circ}$. In a right-angled triangle,$c^2 = a^2 + b^2$. Using the Sine Rule,$a = c \sin A$ and $b = c \sin B$. Substituting t

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$\sin (A-B)$

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(B) Given $\angle C = 90^{\circ}$. In a right-angled triangle,$c^2 = a^2 + b^2$. Using the Sine Rule,$a = c \sin A$ and $b = c \sin B$. Substituting these into the expression: $\frac{a^2-b^2}{a^2+b^2} = \frac{(c \sin A)^2 - (c \sin B)^2}{c^2} = \sin^2 A - \sin^2 B$. Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$. Since $A+B+C = 180^{\circ}$ and $C = 90^{\circ}$,we have $A+B = 90^{\circ}$. Therefore,$\sin(A+B) = \sin 90^{\circ} = 1$. Thus,$\sin^2 A - \sin^2 B = 1 \cdot \sin(A-B) = \sin(A-B)$.

In a $\triangle ABC$,if $\angle C = 90^{\circ}$,then $\frac{a^2-b^2}{a^2+b^2}$ is equal to

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