In ∆ABC, prove that:

Question:

In ∆ABC, prove that:

$a(\cos B+\cos C-1)+b(\cos C+\cos A-1)+c(\cos A+\cos B-1)=0$

Solution:

Consider the LHS of the given equation.

LHS $=a(\cos B+\cos C-1)+b(\cos C+\cos A-1)+c(\cos A+\cos B-1)$

$=a \cos B+b \cos C+a \cos C+b \cos A+c \cos A+c \cos B-(a+b+c)$

$=(a \cos B+b \cos A)+(b \cos C+c \cos B)+(a \cos C+c \cos A)-(a+b+c)$

$=c+a+b-(a+b+c) \quad$. (Using projection formula : $a=b \cos C+c \cos B, b=a \cos C+c \cos A, c=a \cos B+b \cos A$ )

$=0=\mathrm{RHS}$

Hence proved.

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