In any triangle ABC, prove that

c - bcos A / b - ccos A = cos B/ cos C

Urgent

c - bcos A / b - ccos A = cos B/ cos C

By the projection formula:

c = bcosA + acosB  and b = ccosA + acosC

Therefore, c - bcosA = acosB and b - ccosA = acosC

Therefore, (c - bcosA)/(b - cosA) = cosB/cosC