In any triangle A B C , b + c cos A + c + a cos B + a + b cos C equals

Given

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

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

$=\left(b\cos A+a\cos B\right)+\left(c\cos A+a\cos C\right)+\left(c\cos B+b\cos C\right) ...\left(1\right)$

By projection formula, we have

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

Therefore, from $\left(1\right)$, we have

$=a+b+c$