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Question
If cosec A + sec A = cosec B + sec B, prove that tan A tan B = \[\cot\frac{A + B}{2}\].
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Solution
\[\frac{1}{\sin A} + \frac{1}{\cos A} = \frac{1}{\sin B} + \frac{1}{\cos B}\]
\[ \Rightarrow \frac{1}{\sin A} - \frac{1}{\sin B} = \frac{1}{\cos B} - \frac{1}{\cos A}\]
\[ \Rightarrow \frac{\sin B - \sin A}{\sin A\sin B} = \frac{\cos A - \cos B}{\cos A\cos B}\]
\[ \Rightarrow \frac{\sin B - \sin A}{\cos A - \cos B} = \frac{\sin A\sin B}{\cos A\cos B}\]
\[ \Rightarrow \frac{2\sin\left( \frac{B - A}{2} \right)\cos\left( \frac{A + B}{2} \right)}{- 2\sin\left( \frac{A - B}{2} \right)\sin\left( \frac{A + B}{2} \right)} = \frac{\sin A\sin B}{\cos A\cos B}\]
\[ \Rightarrow \frac{- \sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)}{- \sin\left( \frac{A - B}{2} \right)\sin\left( \frac{A + B}{2} \right)} = \frac{\sin A\sin B}{\cos A\cos B}\]
\[ \Rightarrow \frac{\cos\left( \frac{A + B}{2} \right)}{\sin\left( \frac{A + B}{2} \right)} = \frac{\sin A\sin B}{\cos A\cos B}\]
\[ \Rightarrow \cot\left( \frac{A + B}{2} \right) = \tan A\tan B\]
Hence proved.
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