Question

a(cosB+cosC-1)+b(cosC+cosA-1)+c(cosA+cosB-1)=0​

Answer 1

Answer:

Proved!

Step-by-step explanation:

Given Problem:

a(cosB+cosC-1)+b(cosC+cosA-1)+c(cosA+cosB-1) = 0​

Solution:

Let ABC be any triangle.

1) Considerthe L.H.S  (left hand side) of the given equation:

LHS = $a(cosB+cosC-1)+b(cosC+cosA-1)+c(cosA+cosB-1)$

$=AcosB+BcosC+AcosC+BcosA+CcosA+CcosB-(a+b+c)$

$=(acosB+bcosA)+(bcosC+ccosB)+(acosC+ccosA)-(a+b+c)$

$=c+a+b-(a+b+c)$

We know that,

Projection formula:

$\implies\ a=bcosC+ccosB, b=acosC+ccosA, c=acosB+bcosA)$

= 0 =RHS

(2) Consider the LHS of the given equation:

$LHS= \dfrac{CosA}{b\:cosC+c\:cosB}+\dfrac{CosB}{c\:cosA+a\:cosC} +\dfrac{cosC}{a\:cosB+b\:cosA}$

$=\dfrac{cosA}{a} +\dfrac{cosB}{b}+\dfrac{cosC}{c}$

$=\dfrac{b^2+c^2-a^2}{2abc} +\dfrac{c^2+a^2-c^2}{2abc}$

$=\dfrac{a^2+b^2-c^2}{2abc}$

$=\dfrac{b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2}{2abc}$

$=\dfrac{a^2+b^2+c^2}{2abc} = RHS$

Hence Proved!!