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Question
In ∆ABC, prove the following: \[c \left( a \cos B - b \cos A \right) = a^2 - b^2\]
Solution
Consider
\[c\left( a\cos B - b\cos A \right) = ca\cos B - cb\cos A\]
\[ = ca\left( \frac{a^2 + c^2 - b^2}{2ac} \right) - cb\left( \frac{b^2 + c^2 - a^2}{2bc} \right)\]
\[ = \frac{a^2 + c^2 - b^2 - b^2 - c^2 + a^2}{2}\]
\[ = \frac{2\left( a^2 - b^2 \right)}{2}\]
\[ = a^2 - b^2 \]
Hence proved.
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