In any triangle `A B C` , prove that following: `\ (a^2"sin"(B-C))/(sin A)+(b^2"sin"(C-A))/(sin B)+(c^2"sin"(A-B))/(sin C)=0`

Explanation

`(a^(2) sin (B - C))/(sin B + sin C) = (4R^(2) sin ^(2) A sin (B - C))/(sin B + sin C) ( :' a = 2R sin A)`
`= (4R^(2) sin A sin (B + C) sin (B - C))/(sin B + sin C)`
`= (4R^(2) sin A (sin^(2) B - sin^(2) C))/(sin B + sin C)`
`= 4R^(2) sin A (sin B - sin C)`
(i) similarly, `(b^(2) sin (C - A))/(sin C + sin A) = 4R^(2) sin B (sin C - sin A)`
and `(c^(2) sin (A - B))/(sin A + sin B) = 4R^(2) sin C (sin A - sin B)`
Adding we get
`(a^(2) sin (B - C))/(sin B + sin C) + (b^(2) sin (C - A))/(sin C + sin A) + (c^(2) sin (A - B))/(sin A + sin B) = 0`