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Question
In triangle ABC, prove the following:
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Solution
Let
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]
Consider the LHS of he equation
LHS \[= k\sin B\sin B - k\sin C\sin C\]
\[= k\left( \sin^2 B - \sin^2 C \right)\]
\[ = k\left[ \sin\left( B + C \right)\sin\left( B - C \right) \right] \left[ \because \sin^2 B - \sin^2 C = \sin\left( B + C \right)\sin\left( B - C \right) \right]\]
\[ = k\left[ \sin\left( \pi - A \right)\sin\left( B - C \right) \right] \left[ \because A + B + C = \pi \right]\]
\[ = k\sin A\sin\left( B - C \right) \left[ \because a = k\sin A \right]\]
\[ = a\sin\left( B - C \right) = RHS\]
\[\text{ Hence proved }.\]
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