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

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(B) Given the equation $\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = 1$,we have $\sin(A-B) = \frac{a^2-b^2}{a^2+b^2}$.Using the Sine Rule,$a = 2R \sin A$ and $

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$c > a > b$

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(B) Given the equation $\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = 1$,we have $\sin(A-B) = \frac{a^2-b^2}{a^2+b^2}$.Using the Sine Rule,$a = 2R \sin A$ and $b = 2R \sin B$,we get $\frac{a^2-b^2}{a^2+b^2} = \frac{\sin^2 A - \sin^2 B}{\sin^2 A + \sin^2 B} = \frac{\sin(A-B)\sin(A+B)}{\sin^2 A + \sin^2 B}$.Since $\angle C = 90^{\circ}$,$A+B = 90^{\circ}$,so $\sin(A+B) = 1$.Thus,$\sin(A-B) = \frac{\sin(A-B)}{\sin^2 A + \sin^2 B}$.This implies $\sin^2 A + \sin^2 B = 1$. Since $\sin^2 A + \cos^2 A = 1$ and $\sin B = \cos A$,this is consistent.For $\sin(A-B) > 0$,we must have $A > B$,which implies $a > b$.In a right-angled triangle with $\angle C = 90^{\circ}$,the hypotenuse $c$ is the longest side.Therefore,$c > a > b$.

In a $\triangle ABC$,if $\angle C = 90^{\circ}$ and $\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = 1$,then which of the following is true?

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