Question

In any $\Delta ABC,$ prove that

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

Answer 1

By the sine rule, we have

$\frac{a}{sinA}+\frac{b}{sinB}+\frac{c}{sinC}=k\left(say\right)$

$\Rightarrow a=ksinA,b=ksinB,c=ksinC.$

$\therefore RHS=\frac{(b^2-c^2)}{a^2}=\frac{k^{2}si{n}^{2}B-k^{2}si{n}^{2}C}{si{n}^2(B+C)}$

= $\frac{(si{n}^{2}B-si{n}^{2}C)}{si{n}^{2}A}=\frac{sin(B+C).sin(B-C)}{si{n}^2(B+C)}$

$\left[\begin{array}{c}\because (A+B+C)=\pi \Rightarrow A=\pi -(B+C)\\ \therefore sinA=sin|\pi -(B+C)|=sin(B-C)\end{array}\right]$

= $\frac{sin(B-C)}{sin(B+C)}=LHS$

$\therefore$ RHS = LHS

Hence, $\frac{sin(B-C)}{sin(B+C)}=sin\frac{(A-B)}{2}$