With usual notations,if triangle ABC is right | vedclass

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(B) Given that in $	riangle ABC, \angle C = 90^{\circ}$,so $A+B = 90^{\circ} \Rightarrow B = 90^{\circ}-A$. Using the sine rule,$a = k \sin A$ and $b

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(B) Given that in $	riangle ABC, \angle C = 90^{\circ}$,so $A+B = 90^{\circ} \Rightarrow B = 90^{\circ}-A$. Using the sine rule,$a = k \sin A$ and $b = k \sin B$. Substituting these into the expression: $\frac{a^{2}+b^{2}}{a^{2}-b^{2}} \sin (A-B) = \frac{\sin^{2} A + \sin^{2} B}{\sin^{2} A - \sin^{2} B} \sin (A-B)$. Since $B = 90^{\circ}-A$,$\sin B = \cos A$ and $\cos B = \sin A$. Thus,$\sin^{2} A + \sin^{2} B = \sin^{2} A + \cos^{2} A = 1$. And $\sin^{2} A - \sin^{2} B = \sin^{2} A - \cos^{2} A = -\cos 2A$. Also,$\sin (A-B) = \sin (A - (90^{\circ}-A)) = \sin (2A - 90^{\circ}) = -\cos 2A$. Substituting these values: $\frac{1}{-\cos 2A} \cdot (-\cos 2A) = 1$.

With usual notations,if triangle $ABC$ is right-angled at $C$,then $\left(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\right) \sin (A-B) =$

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