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Question
Prove that:
\[\frac{\tan \left( A + B \right)}{\cot \left( A - B \right)} = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}\]
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Solution
\[\text{ LHS }= \frac{\tan\left( A + B \right)}{cot\left( A - B \right)}\]
\[ = \frac{\tan\left( A + B \right)}{\frac{1}{\tan\left( A - B \right)}}\]
\[ = \tan\left( A + B \right) \times \tan\left( A - B \right)\]
\[ = \frac{\tan A + \tan B}{1 - \tan A \tan B} \times \frac{\tan A - \tan B}{1 + \tan A\tan B}\]
\[ = \frac{\left( \tan A + \tan B \right)\left( \tan A - \tan B \right)}{\left( 1 - \tan A \tan B \right)\left( 1 + \tan A \tan B \right)}\]
\[ = \frac{\left( \tan A \right)^2 - \left( \tan B \right)^2}{\left( 1 \right)^2 - \left( \tan A \tan B \right)^2}\]
\[ = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}\]
= RHS
Hence proved.
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