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प्रश्न
In ∆ABC, prove that \[a \left( \cos B + \cos C - 1 \right) + b \left( \cos C + \cos A - 1 \right) + c\left( \cos A + \cos B - 1 \right) = 0\]
उत्तर
\[\text{ Consider the LHS of the given equation } . \]
\[LHS = a\left( \cos B + \cos C - 1 \right) + b\left( \cos C + \cos A - 1 \right) + c\left( \cos A + \cos B - 1 \right)\]
\[ = a\cos B + b\cos C + a\cos C + b\cos A + c\cos A + c\cos B - \left( a + b + c \right)\]
\[ = \left( a\cos B + b\cos A \right) + \left( b\cos C + c\cos B \right) + \left( a\cos C + c\cos A \right) - \left( a + b + c \right) \]
\[ = c + a + b - \left( a + b + c \right) . . \left( \text{ Using projection formula }: a = b\cos C + c\cos B, b = a\cos C + c\cos A, c = a\cos B + b\cos A \right)\]
\[ = 0 = RHS \]
\[\text{ Hence proved }.\]
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