If A, B, C, D be the angles of a quadrilateral, then $\frac{\tan \mathrm{A}+\tan \mathrm{B}+\tan \mathrm{C}+\tan \mathrm{D}}{\cot \mathrm{A}+\cot \mathrm{B}+\cot \mathrm{C}+\cot \mathrm{D}}=$

$\begin{array}{l}\tan \left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)=\tan \left(2\mathrm{\pi }-D\right)\\ \frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}=-\tan D\\ \tan A+\tan B+\tan C-\tan A\tan B\tan C=-\tan D\left(1-\tan A\tan B-\tan B\tan C-\tan C\tan A\right)\\ \end{array}$

$\tan A+\tan B+\tan C+\tan D=\tan A\tan B\tan C\tan D(cotA+cotB+cotC+cotD)$