Question

For any triangle $\Delta ABC$ prove that $\frac{\sin \left(B-C\right)}{\sin \left(B+C\right)}=\frac{b^2-c^2}{a^2}$

Answer 1

To Prove : $\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^2-c^2}{a^2}$

L,H,S

$\Rightarrow \frac{\sin B\cos C-\cos B\sin C}{\sin (180-A)}$

$\Rightarrow \frac{\sin B\cos C-\cos B\sin C}{\sin A}$

$\Rightarrow \frac{\left(bk\right)\frac{a^2+b^2-c^2}{2ab}-\left(\frac{c^2+a^2-b^2}{2ac}\right)ck}{ak}$ [sine rule and cosine rule ]

$\Rightarrow \frac{k}{2ak}\left\{\frac{a^2+b^2-c^2-c^2-a^2+b^2}{a}\right\}$

$\Rightarrow \frac{2(b^2-c^2)}{2a^2}=\frac{b^2-c^2}{a^2}=$R.H.S

Since L.H.S=R.H.S

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

Hence Proved