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Question
Prove that:
cos2 A + cos2 B − 2 cos A cos B cos (A + B) = sin2 (A + B)
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Solution
\[\text{ LHS }= \cos^2 A + \cos^2 B - 2\cos A \cos B \cos\left( A + B \right)\]
\[ = \cos^2 A + 1 - \sin^2 B - 2\cos A \cos B \cos\left( A + B \right)\]
\[ = 1 + \cos^2 A - \sin^2 B - 2\cos A \cos B \cos\left( A + B \right)\]
\[ = 1 + \cos^2 A - \sin^2 B - 2\cos A \cos B \cos\left( A + B \right)\]
\[ = 1 + \cos\left( A + B \right)\cos\left( A - B \right) - 2\cos A \cos B \cos\left( A + B \right)\]
\[ = 1 + \cos\left( A + B \right)\left\{ \cos\left( A - B \right) - 2\cos A \cos B \right\}\]
\[ = 1 + \cos\left( A + B \right)\left( \cos A \cos B + \sin A \sin B - 2\cos A \cos B \right)\]
\[ = 1 + \cos\left( A + B \right)\left( - \cos A \cos B + \sin A \sin B \right)\]
\[ = 1 - \cos\left( A + B \right)\left( \cos A \cos B - \sin A \sin B \right)\]
\[ = 1 - \cos\left( A + B \right)\cos\left( A + B \right)\]
\[ = 1 - \cos^2 \left( A + B \right)\]
\[ = \sin^2 \left( A + B \right)\]
= RHS
Hence proved.
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