If $\operatorname{cosec} A+\sec A=\operatorname{cosec} B+\sec B$, prove that $\tan A \tan B=\cot \frac{A+B}{2}$.
Given:
$\frac{1}{\sin A}+\frac{1}{\cos A}=\frac{1}{\sin B}+\frac{1}{\cos B}$
$\Rightarrow \frac{1}{\sin A}-\frac{1}{\sin B}=\frac{1}{\cos B}-\frac{1}{\cos A}$
$\Rightarrow \frac{\sin B-\sin A}{\sin A \sin B}=\frac{\cos A-\cos B}{\cos A \cos B}$
$\Rightarrow \frac{\sin B-\sin A}{\cos A-\cos B}=\frac{\sin A \sin B}{\cos A \cos B}$
$\Rightarrow \frac{2 \sin \left(\frac{B-A}{2}\right) \cos \left(\frac{A+B}{2}\right)}{-2 \sin \left(\frac{A-B}{2}\right) \sin \left(\frac{A+B}{2}\right)}=\frac{\sin A \sin B}{\cos A \cos B}$
$\Rightarrow \frac{-\sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)}{-\sin \left(\frac{A-B}{2}\right) \sin \left(\frac{A+B}{2}\right)}=\frac{\sin A \sin B}{\cos A \cos B}$
$\Rightarrow \frac{\cos \left(\frac{A+B}{2}\right)}{\sin \left(\frac{A+B}{2}\right)}=\frac{\sin A \sin B}{\cos A \cos B}$
$\Rightarrow \cot \left(\frac{A+B}{2}\right)=\tan A \tan B$
Hence proed.