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Question
In ∆ABC, prove the following: \[b \left( c \cos A - a \cos C \right) = c^2 - a^2\]
Solution
Let ABC be any triangle.
\[\text{ Consider }\]
\[ b\left( c\cos A - a\cos C \right) = bc\cos A - ab\cos C\]
\[ = bc\left( \frac{b^2 + c^2 - a^2}{2bc} \right) - ab\left( \frac{a^2 + b^2 - c^2}{2ab} \right) \]
\[ = \frac{b^2 + c^2 - a^2 - a^2 - b^2 + c^2}{2}\]
\[ = \frac{2\left( c^2 - a^2 \right)}{2}\]
\[ = c^2 - a^2\]
Hence proved.
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