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Question
If tan A + tan B = a and cot A + cot B = b, prove that cot (A + B) \[\frac{1}{a} - \frac{1}{b}\].
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Solution
Given:
\[\cot A + \cot B = b\]
\[ \Rightarrow \frac{1}{\tan A} + \frac{1}{\tan B} = b\]
\[ \Rightarrow \frac{\tan A + \tan B}{\tan A\tan B} = b\]
Now,
\[\text{ RHS }= \frac{1}{a} - \frac{1}{b} \]
\[ = \frac{1}{\tan A + \tan B} - \frac{\tan A \tan B}{\tan A + tan B}\]
\[ = \frac{1 - \tan A \tan B}{\tan A + \tan B} \]
\[ = \cot (A + B) \]
= LHS
Hence proved .
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