a( + -1)+b( + -1) +c( + -1)=0 - Vidyadip

Question

$\text{a}(\cos\text{B}+\cos\text{C}-1)+\text{b}(\cos\text{C}+\cos\text{A}-1)\\+\text{c}(\cos\text{A}+\cos\text{B}-1)=0$

✓

Answer

In any $\triangle\text{ABC},$ we have
$\text{a = b}\cos\text{C + c}\cos\text{B}$
$\text{b = c}\cos\text{A + a}\cos\text{C}$
$\text{c = a}\cos\text{B + b}\cos\text{A}$
Therefore,
$\text{LHS}=\text{a}(\cos\text{B}+\cos\text{C}-1)+\text{b}(\cos\text{C}+\cos\text{A}-1)\\+\text{c}(\cos\text{A}+\cos\text{B}-1)$
$=\text{a}\cos\text{B + a}\cos\text{C}-\text{a + b}\cos\text{C + b}\cos\text{A}-\text{b + c}\cos\text{A + c}\cos\text{B}-\text{c}$
$=\text{c}-\text{b}\cos\text{A + a}\cos\text{C}-\text{a + a}-\text{c}\cos\text{B}\\+\text{b}\cos\text{A}-\text{b + b}-\text{a}\cos\text{C + c}\cos\text{B}-\text{c}$
$=0=\text{RHS}$
Hence proved.

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