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Question
\[a \left( \cos B \cos C + \cos A \right) = b \left( \cos C \cos A + \cos B \right) = c \left( \cos A \cos B + \cos C \right)\]
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Solution
Suppose \[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]
Consider:
\[a\left( \cos B\cos C + \cos A \right)\]
\[ = k\sin A\left( \cos B\cos C + \cos A \right) \]
\[ = k\left( \sin A\cos B\cos C + \cos A\sin A \right)\]
\[ = k\left[ \frac{1}{2}\cos C\left\{ \sin\left( A + B \right) + \sin\left( A - B \right) \right\} + \sin A\cos A \right]\]
\[ = k\left[ \frac{1}{2}\left\{ \sin\left( A + B \right)\cos C + \sin\left( A - B \right)\cos C \right\} + \sin A\cos A \right]\]
\[ = k\left[ \frac{1}{2}\left\{ \frac{1}{2}\left[ \sin\left( A + B + C \right) + \sin\left( A + B - C \right) + \sin\left( A - B + C \right) + \sin\left( A - B - C \right) \right] \right\} + \sin A\cos A \right]\]
\[ = k\left[ \frac{1}{4}\left\{ sin\pi + \sin\left( \pi - 2C \right) + \sin\left( \pi - 2B \right) - \sin\left( \pi - 2A \right) \right\} + \frac{\sin2A}{2} \right] \left( \because A + B + C = \pi \right)\]
\[ = \frac{k}{4}\left( \sin2C + \sin2B + \sin2A \right) . . . . \left( 1 \right)\]
\[\text{ and } \]
\[b\left( \cos A\cos C + \cos B \right)\]
\[ = k\left( \sin B\cos A\cos C + sinBcosB \right)\]
\[ = k\left[ \frac{1}{2}\cos A\left\{ \sin\left( B + C \right) + \sin\left( B - C \right) \right\} + \frac{\sin2B}{2} \right]\]
\[ = k\left( \frac{1}{2}\left( \sin\left( B + C \right)\cos A + \sin\left( B - C \right)\cos A \right) + \frac{\sin2B}{2} \right)\]
\[ = k\left( \frac{1}{4}\left( \sin\left( B + C + A \right) + \sin\left( B + C - A \right) + \sin\left( B - C + A \right) + \sin\left( B - C - A \right) \right) + \frac{\sin2B}{2} \right)\]
\[ = \frac{k}{4}\left( sin\pi + \sin\left( \pi - 2A \right) + \sin\left( \pi - 2C \right) - \sin\left( \pi - 2B \right) + \frac{\sin2B}{2} \right) \left( \because A + B + C = \pi \right)\]
\[ = \frac{k}{4}\left( \sin2A + \sin2C + \sin2B \right) . . . \left( 2 \right)\]
\[\text{ Similarly }, \]
\[c\left( \cos A\cos B + \cos C \right) = \frac{k}{4}\left( \sin2A + \sin2B + \sin2C \right) . . . \left( 3 \right)\]
From (1), (2) and (3), we get:
\[a \left( \cos B \cos C + \cos A \right) = b \left( \cos C \cos A + \cos B \right) = c \left( \cos A \cos B + \cos C \right)\]
Hence proved.
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