Question

Prove that $\frac{a^2\sin (B-C)}{\sin B+\sin C}+\frac{b^2\sin (C-A)}{\sin C+\sin A}+\frac{c^2\sin (A-B)}{\sin A+\sin B}=0$

Answer 1

Given:

We know that,

$a=2RsinA$ and $A+B+C=\pi$

$\therefore$ $A=\pi -(B+C)$

$\therefore$ $sinA=sin(\pi -(B+C\left)\right)=sin(B+C)$

$\frac{a^2\sin (B-C)}{\sin B+\sin C}=\frac{4R^2{\sin }^{2}A\sin (B-C)}{\sin B+\sin C}$ (using the formula in triangle)

$=\frac{4R^2\sin A\sin (B+C)\sin (B-C)}{\sin B+\sin C}$

$=\frac{4R^2\sin A({\sin }^{2}B-{\sin }^{2}C)}{\sin B+\sin C}$

$=4R^2\sin A(\sin B-\sin C)$

Similarly, $\frac{b^2\sin (C-A)}{\sin C+\sin A}=4R^2\sin B(\sin C-\sin A)$

and

$\frac{c^2\sin (A-B)}{\sin A+\sin B}=4R^2\sin C(\sin A-\sin B)$

Adding we get $\frac{a^2\sin (B-C)}{\sin B+\sin C}+\frac{b^2\sin (C-A)}{\sin C+\sin A}+\frac{c^2\sin (A-B)}{\sin A+\sin B}=0$