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Question
In a ∆ABC, prove that:
cos (A + B) + cos C = 0
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Solution
In ∆ ABC:
\[A + B + C = \pi\]
\[ \therefore A + B = \pi - C\]
\[\text{ Now, LHS }= \cos\left( A + B \right) + \cos C\]
\[ = \cos\left( \pi - C \right) + \cos C\]
\[ = - \cos\left( C \right) + \cos C \left[ \because \cos\left( \pi - C \right) = - \cos\left( C \right) \right] \]
\[ = 0\]
= RHS
Hence proved .
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