Question

If $△ABC$ is such that $\angle A={90}^o,\angle B=\angle C$, then $\frac{b^2+c^2}{b^2-c^2}\sin (B-C)=$

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $1$
• D. $\frac{3}{2}$

Answer 1

Answer: (C)

$1$

By Sine rule, $a\propto \sin A,b\propto \sin B,c\propto \sin C$

$\frac{b^2+c^2}{b^2-c^2}\sin (B-C)=\frac{{\sin }^{2}B+{\sin }^{2}C}{{\sin }^{2}B-{\sin }^{2}C}\times \sin (B-C)$

$\frac{b^2+c^2}{b^2-c^2}\sin (B-C)=\frac{{\sin }^{2}B+{\sin }^{2}C}{\sin (B+C)\sin (B-C)}\times \sin (B-C)$

Since $\angle A={90}^{\circ }$ and $\angle B=\angle C$, we can say that $\angle B=\angle C={45}^{\circ }$.

$\frac{b^2+c^2}{b^2-c^2}\sin (B-C)=\frac{{\sin }^{2}45+{\sin }^{2}45}{\sin 90}=1$