In ∆ABC, prove that:
$a (\cos B + \cos C - 1) + b (\cos C + \cos A - 1) + c (\cos A + \cos B - 1) = 0$

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Solution

$Consider the LHS of the given equation . LHS = a (\cos B + \cos C - 1) + b (\cos C + \cos A - 1) + c (\cos A + \cos B - 1) = a \cos B + b \cos C + a \cos C + b \cos A + c \cos A + c \cos B - (a + b + c) = (a \cos B + b \cos A) + (b \cos C + c \cos B) + (a \cos C + c \cos A) - (a + b + c) = c + a + b - (a + b + c) . . (Using projection formula : a = b \cos C + c \cos B, b = a \cos C + c \cos A, c = a \cos B + b \cos A) = 0 = RHS Hence proved .$

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In ∆ABC, prove that: a cos B+cos C-1+b cos C+cos A-1+ccos A+cos B-1=0

Consider the LHS of the given equation.LHS=acosB+cosC-1+bcosC+cosA-1+ccosA+cosB-1 =acosB+bcosC+acosC+bcosA+ccosA+ccosB-a+b+c =acosB+bcosA+bcosC+ccosB+acosC+ccosA-a+b+c =c+a+b-a+b+c ..Using projection formula : a=bcosC+ccosB, b=acosC+ccosA, c=acosB+bcosA =0=RHS Hence proved.

In ∆ABC, prove that:
$a (\cos B + \cos C - 1) + b (\cos C + \cos A - 1) + c (\cos A + \cos B - 1) = 0$