In a ∆ABC, if

In a ∆ABC,

Given $\frac{\sin (A-B)}{\sin (A+B)}=\frac{\sin A \cos B-\cos A \sin B}{\sin A \cos B+\cos A \sin B}$

Since $\quad \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$

$\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$

and $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=x$ say

i. e $\frac{\sin (A-B)}{\sin (A+B)}=\frac{\left(\frac{a}{x}\right)\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)-\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)\left(\frac{b}{x}\right)}{\frac{a}{x}\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)+\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)\left(\frac{b}{x}\right)}$

$=\frac{\frac{a^{2}+c^{2}-b^{2}}{2 c x}-\frac{b^{2}+c^{2}-a^{2}}{2 c x}}{\frac{a^{2}+c^{2}-b^{2}}{3 x x}+\frac{b^{2}+c^{2}-a^{2}}{2 x z}}$

$=\frac{a^{2}+c^{2}-b^{2}-b^{2}-c^{2}+a^{2}}{a^{2}+c^{2}-b^{2}+b^{2}+c^{2}-a^{2}}$

$=\frac{2 a^{2}-2 b^{2}}{2 c^{2}}=\frac{a^{2}-b^{2}}{c^{2}}=\frac{a^{2}-b^{2}}{k} \quad$ (given)

i. e. $k=c^{2}$