Question

Find the value of : $\frac{a^2\sin (B-C)}{\sin B+\sin C}+\frac{b^2\sin (C-A)}{\sin C+\sin A}+\frac{c^2\sin (A-B)}{\sin A+\sin B}$

Answer 1

From the properties of triangle.

$a=ksinA$

$b=ksinB$

$c=ksinC$

Substitute in the above equation.

$\frac{k^2{sin}^{2}Asin(B-C)}{(sinB+sinC)}+\frac{k^2{sin}^{2}Bsin(C-A)}{(sinC+sinA)}+\frac{k^2{sin}^{2}Csin(A-B)}{(sinA+sinB)}...\left(i\right)$

Hence the above equation can be written as,

$\frac{k^{2}sinAsin(B+C)sin(B-C)}{(sinB+sinC)}+\frac{k^{2}sinBsin(C+A)sin(C-A)}{(sinC+sinA)}+\frac{k^{2}sinCsin(A+B)sin(A-B)}{(sinA+sinB)}=\frac{k^{2}sinA({sin}^{2}B-{sin}^{2}C)}{(sinB+sinC)}+\frac{k^{2}sinB({sin}^{2}C-{sin}^{2}A)}{(sinC+sinA)}+\frac{k^{2}sinC({sin}^{2}A-{sin}^{2}B)}{(sinA+sinB)}=k^2\left(sinA\right(sinB-sinC)+sinB(sinC-sinA)+sinC(sinA-sinB\left)\right)=0$