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(D) Given $\angle C = 90^\circ$, so $A + B = 90^\circ$, which implies $B = 90^\circ - A$.Using the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \

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

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(D) Given $\angle C = 90^\circ$, so $A + B = 90^\circ$, which implies $B = 90^\circ - A$.Using the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.Thus, $a = 2R \sin A$ and $b = 2R \sin B = 2R \sin(90^\circ - A) = 2R \cos A$.Now, consider $\frac{a^2 - b^2}{a^2 + b^2} = \frac{(2R \sin A)^2 - (2R \cos A)^2}{(2R \sin A)^2 + (2R \cos A)^2}$.$= \frac{4R^2(\sin^2 A - \cos^2 A)}{4R^2(\sin^2 A + \cos^2 A)} = \frac{-(\cos^2 A - \sin^2 A)}{1} = -\cos(2A)$.Since $B = 90^\circ - A$, then $A - B = A - (90^\circ - A) = 2A - 90^\circ$.Therefore, $\sin(A - B) = \sin(2A - 90^\circ) = -\sin(90^\circ - 2A) = -\cos(2A)$.Comparing the two results, $\sin(A - B) = \frac{a^2 - b^2}{a^2 + b^2}$.

In $\triangle ABC$, with the usual notations, $\angle C = 90^\circ$, then $\sin(A - B)$ is equal to....

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