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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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