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Question
In triangle ABC, prove the following:
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Solution
\[Let\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k . . . \left( 1 \right)\]
\[\text{ Consider the LHS of the equation a }\cos A + b\cos B + c \cos C . \]
\[a\cos A + b\cos B + c\cos C = k\left( \sin A\cos A + \sin B\cos B + \sin C\cos C \right) \]
\[ = \frac{k}{2}\left( 2sinAcosA + 2sinAcosA + 2sinCcosC \right)\]
\[ = \frac{k}{2}\left( \sin2A + \sin2 B + \sin2 C \right)\]
\[ = \frac{k}{2}\left[ 2\sin\left( A + B \right)\cos\left( A - B \right) + 2\sin C\cos C \right]\]
\[ = \frac{k}{2}\left[ 2\sin\left( \pi - C \right)\cos\left( A - B \right) + 2\sin C\cos C \right]\]
\[ = \frac{k}{2}\left[ 2\sin C\cos\left( A - B \right) + 2\sin C\cos C \right]\]
\[ = \frac{2k\sin C}{2}\left[ \cos\left( A - B \right) + \cos C \right] \]
\[= k\sin C\left[ \cos\left( A - B \right) + \cos\left\{ \pi - \left( A + B \right) \right\} \right]\]
\[ = k\sin C\left[ \cos\left( A - B \right) - \cos\left( A + B \right) \right]\]
\[ = k\sin C\left[ 2\sin Asin B \right]\]
\[ = 2k\sin A\sin B\sin C . . . (1)\]
\[\text{ Now }, \]
\[\text{ on putting } k\sin C = \text{ C in equation } (1), \text{ we get }: \]
\[2c\sin A\sin B\]
\[\text{ and on putting k }\sin B = \text{ b in equation } (1), \text{ we get }: \]
\[2b\sin A\sin C\]
So, from (1), we have
Hence proved.
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