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In ∆Abc, Prove that a ( Cos B + Cos C − 1 ) + B ( Cos C + Cos a − 1 ) + C ( Cos a + Cos B − 1 ) =

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Question

In ∆ABC, prove that  \[a \left( \cos B + \cos C - 1 \right) + b \left( \cos C + \cos A - 1 \right) + c\left( \cos A + \cos B - 1 \right) = 0\]

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Solution

\[\text{ Consider the LHS of the given equation } . \]

\[LHS = a\left( \cos B + \cos C - 1 \right) + b\left( \cos C + \cos A - 1 \right) + c\left( \cos A + \cos B - 1 \right)\]

\[ = a\cos B + b\cos C + a\cos C + b\cos A + c\cos A + c\cos B - \left( a + b + c \right)\]

\[ = \left( a\cos B + b\cos A \right) + \left( b\cos C + c\cos B \right) + \left( a\cos C + c\cos A \right) - \left( a + b + c \right) \]

\[ = c + a + b - \left( a + b + c \right) . . \left( \text{ Using projection formula }: a = b\cos C + c\cos B, b = a\cos C + c\cos A, c = a\cos B + b\cos A \right)\]

\[ = 0 = RHS \]

\[\text{ Hence proved }.\]

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Sine and Cosine Formulae and Their Applications
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