Question

In any $\triangle A B C, \frac{\left(b^2-c^2\right)}{a^2}$ is equal to

• A. $\frac{\sin (B+C)}{\sin (B-C)}$
• B. $\frac{\sin (B-C)}{\sin (B+C)}$
• C. $\sin (B+C)$
• D. $\sin (B-C)$

Answer 1

Answer: (B)

$ \because \frac{\left(b^2-c^2\right)}{a^2} =\frac{K^2 \sin ^2 B-K^2 \sin ^2 C}{K^2 \sin ^2 A}$
$= \frac{\left(\sin ^2 B-\sin ^2 C\right)}{\sin ^2 A}=\frac{\sin (B+C) \cdot \sin (B-C)}{\sin ^2(B+C)} $
$\begin{Bmatrix}\because & (A+B+C)=\pi \Rightarrow A=\pi-(B+C) \\ \therefore & \sin A=\sin [\pi-(B+C)]=\sin (B+C)\end{Bmatrix}$
$ =\frac{\sin (B-C)}{\sin (B+C)} $
Hence, $ \frac{\sin (B-C)}{\sin (B+C)}=\frac{\left(b^2-c^2\right)}{a^2}$