Question

In a triangle ABC, prove that $\left(a-b\right)\left(1+\cos C\right)=c\left(\cos B-\cos A\right).$

Answer 1

To prove:- $(a-b)(1+\cos C)=c(\cos B-\cos A)$

Proof:- In $△ABC$, by projection law

$a=b\cos C+c\cos B$

$b=a\cos C+c\cos A$

$c=b\cos A+a\cos B$

LHS:-

$(a-b)(1+\cos C)$

$=(a+a\cos C-b-b\cos C)$ (Substituting the value of a,b from above)

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

$=c(\cos B-\cos A)$

=RHS

Hence, proved.