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If $\triangle A B C$ is such that $\angle A=90^{\circ}, \angle B \neq \angle \mathrm{C}$, then $\frac{b^2+c^2}{b^2-c^2} \sin (B-C)$ is equal to
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Given, $\angle A=90^{\circ}, \angle B \neq \angle C$
$$
\begin{aligned}
\because \quad & \frac{b^2+c^2}{b^2-c^2} \sin (B-C) \\
= & \frac{k^2 \sin ^2 B+k^2 \sin ^2 C}{k^2 \sin ^2 B-k^2 \sin ^2 C} \sin (B-C) \\
& {\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\right] } \\
= & \frac{\sin ^2 B+\sin ^2 C}{\sin ^2 B-\sin ^2 C} \sin (B-C) \\
= & \frac{\sin ^2 B+\sin ^2\left(90^{\circ}-B\right)}{\sin (B+C) \sin (B-C)} \sin (B-C) \\
= & \frac{\sin ^2 B+\cos ^2 B}{\sin (B+C)}=\frac{1}{\sin 90^{\circ}}=1
\end{aligned}
$$
$$
\begin{aligned}
\because \quad & \frac{b^2+c^2}{b^2-c^2} \sin (B-C) \\
= & \frac{k^2 \sin ^2 B+k^2 \sin ^2 C}{k^2 \sin ^2 B-k^2 \sin ^2 C} \sin (B-C) \\
& {\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\right] } \\
= & \frac{\sin ^2 B+\sin ^2 C}{\sin ^2 B-\sin ^2 C} \sin (B-C) \\
= & \frac{\sin ^2 B+\sin ^2\left(90^{\circ}-B\right)}{\sin (B+C) \sin (B-C)} \sin (B-C) \\
= & \frac{\sin ^2 B+\cos ^2 B}{\sin (B+C)}=\frac{1}{\sin 90^{\circ}}=1
\end{aligned}
$$
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