Question

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

Answer 1

In a triangle $ABC$ we have, $a\cos B+b\cos A=c$.....(1), $b\cos C+c\cos B=a$.....(2) and $a\cos C+c\cos A=b$.....(3).

Now,

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

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

$=c+a+b$. [ Using (1),(2) and (3)]