With usual notations,in triangle ABC,angle C= | vedclass

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(B) Given $\angle C=90^{\circ}$,we have $\angle A+\angle B=90^{\circ}$.Using the sine rule,$a=k \sin A$ and $b=k \sin B$,where $k=2R$.Thus,$\sin A = \

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$\frac{a^2-b^2}{a^2+b^2}$

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(B) Given $\angle C=90^{\circ}$,we have $\angle A+\angle B=90^{\circ}$.Using the sine rule,$a=k \sin A$ and $b=k \sin B$,where $k=2R$.Thus,$\sin A = \frac{a}{k}$ and $\sin B = \frac{b}{k}$.We know that $\sin (A-B) = \sin A \cos B - \cos A \sin B$.Since $\angle B = 90^{\circ}-A$,$\cos B = \sin A$ and $\cos A = \sin B$.Therefore,$\sin (A-B) = \sin A \sin A - \sin B \sin B = \sin^2 A - \sin^2 B$.Substituting $\sin A = \frac{a}{c}$ and $\sin B = \frac{b}{c}$ (since $\sin C = \sin 90^{\circ} = 1$ and $c^2=a^2+b^2$):$\sin (A-B) = (\frac{a}{c})^2 - (\frac{b}{c})^2 = \frac{a^2-b^2}{c^2} = \frac{a^2-b^2}{a^2+b^2}$.

With usual notations,in $\triangle ABC$,$\angle C=90^{\circ}$,then the value of $\sin (A-B)$ is

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