Proof This you can use sine rule and cos rule.
Can You give me a hint to simplify this Problem>>><<<<<<
Proof This a(cosB+cosC-1)+b(cosC+cosA-1)+c(cosA+cosB-1)=0 you can use sine rule and cos rule.
What the question doesn't mention is that ABC is a triangle.
I only used the cosine rule. I converted CosA, cosB and cos c for its cos rule equivant.
then I expanded each bracket.
then I collected terms with a common denominator and simplified the numerators.
then it all just fell out.
Just use sine rule and you will get sinerule : a/sinA = b/sin B = c/sin C = K ( Constant )
LHS = K [ sin A ( cos B + cos C - 1 ) + sin B ( cos C + cos A - 1 ) + sin C ( cos A + cos B - 1 )]
= K [ sin A cos B + sin A cos C - sin A + sin B cos C + sin B cos A - sin B + sin C cos A + sin C cos B - sin C] Rearrange terms
= K [ sin A cos B + sin B cos A + sin A cos C + sin C cos A + sin B cos C + sin C cos B - sin A - sin B - sin C]
= K [ sin ( A + B ) + sin ( A + C ) + sin ( B + C ) - sin A - sin B - sin C]
Now we know that A + B + C = 2 pi OR 180 degree thus we will have sin ( A + B ) = sin C etc Thus we get
LHS = K [ sin C + sin B + sin A - sin A - sin B - sin C ] = 0 = RHS