Question

for triangle ABC prove that : (a+b)cosc + (c+a)cosa (c+a)cosb=a+b+c

Answer 1

(a+b)cosC +(c+a)cosB+(c+a)cosC = (acosB +bcosA)+(acosC+ccosA)+(bcosC+ccosB)= a+b+c

projection formulas of properties of traingle :

acosB+bcosA =c

acosC+ccosA= b

bcosC+ccosB=a

Answer 2

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

Step-by-step explanation:

L.H.S.=$(a+b)\cos C+(b+c)\cos A+(c+a)\cos B$

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

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

Using projection formula,

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

= a + b + c

= R.H.S, proved.

Hence, $(a+b)\cos C+(b+c)\cos A+(c+a)\cos B=a+b+c$, proved.