In triangle ABC, (a+b)cosC + (b+c)cosA + (c+a)cosB =
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Answer:
(b+c)cosA+(c+a)cosB+(a+b)cosC
→ bcosA+ccosA+ccosB+acosB+acosC+bcosC
→ (bcosC+ccosB)+(ccosA+acosC)+(acosB+bcosA) ----( 1 )
Using projection formula,
a=(bcosC+ccosB)
b=(ccosA+acosC)
c=(acosB+bcosA)
Substituting above values in ( 1 ) we get,
→a+b+c
∴ (b+c)cosA+(c+a)cosB+(a+b)cosC=a+b+c
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