If triangle ABC is such that angle A=90^{circ | vedclass

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(C) Given,$\angle A=90^{\circ}$ and $\angle B 
eq \angle C$. Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,we have

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(C) Given,$\angle A=90^{\circ}$ and $\angle B 
eq \angle C$. Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,we have $b = k \sin B$ and $c = k \sin C$. Substituting these into the expression: $\frac{b^2+c^2}{b^2-c^2} \sin (B-C) = \frac{k^2 \sin^2 B + k^2 \sin^2 C}{k^2 \sin^2 B - k^2 \sin^2 C} \sin (B-C)$ $= \frac{\sin^2 B + \sin^2 C}{\sin^2 B - \sin^2 C} \sin (B-C)$ Since $\angle A = 90^{\circ}$,then $B+C = 90^{\circ}$,so $C = 90^{\circ}-B$. Thus,$\sin C = \cos B$ and $\sin^2 C = \cos^2 B$. Also,$\sin^2 B - \sin^2 C = \sin(B+C) \sin(B-C) = \sin(90^{\circ}) \sin(B-C) = 1 \cdot \sin(B-C)$. Substituting these: $= \frac{\sin^2 B + \cos^2 B}{\sin(B+C)} \cdot \sin(B-C) = \frac{1}{\sin(90^{\circ})} = 1$.

If $\triangle ABC$ is such that $\angle A=90^{\circ}$ and $\angle B \neq \angle C$,then $\frac{b^2+c^2}{b^2-c^2} \sin (B-C)$ is equal to

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