In any `DeltaABC`, prove that `"a sin (B - C) + bsin(C - A) + csin(A - B)=0`.

In any `DeltaABC`, prove that `"a sin (B - C) + bsin(C - A) + csin(A -

1.	In any `DeltaABC`, prove that `"a sin (B - C) + bsin(C - A) + csin(A - B)=0`.
Answer» By the sine rule, we have `a/("sin A")=b/("sin B")=c/("sin C")` `rArr" "("sin A")/a=("sin ")/b=("sin C")/c="k (say)"` `rArr" sin A = ka, sin B = kb and sin C = kc"`. Substituting the values of sin B and C and using cosine formulae, we have `"a sin (B - C) = a(sin B cos C - cos B sin C)"` `=a["kb"((a^(2)+b^(2)-c^(2)))/(2ab)"-kc"((c^(2)+a^(2)-b^(2)))/(2ac)]` `=k/2[(a^(2)+b^(2)-c^(2))-(c^(2)+a^(2)-b^(2))=k(b^(2)-c^(2))]`. Similarly, b `"sin(C - A)=b(sin C cos A - cos C sin A)"=k(c^(2)-a^(2))`. `:." a sin(B - C)+b sin(C - A) + csin(A - B)"` `=k(b^(2)-c^(2))+k(c^(2)+a^(2))+k(a^(2)-b^(2))` `=k(b^(2)-c^(2)-a^(2)+a^(2)-b^(2))=kxx0=0`.