Problem
Let $ABCD$ be a convex quadrilateral with no right angles. Show that $$ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot D. $$
Source: Geometry Unbound by Kiran Kedlaya.
Attempt: Well, all we really know about a convex quadrilateral are that $ \angle A + \angle B + \angle C + \angle D = 360^\circ $ and that the polygon is convex. Well, that's obvious. But any starts to a proof of this useful fact would be helpful!
Multiply the original identity by $\tan A \tan B \tan C \tan D$ $$ \tan A + \tan B + \tan C + \tan D = \tan B \tan C \tan D + \tan A \tan C \tan D + \tan A \tan B \tan D + \tan A \tan B \tan C. $$ then find $\tan D$ from it $$ \tan D = \frac{\tan A \tan B \tan C - \tan A - \tan B - \tan C}{1 - \tan B \tan C - \tan A \tan C - \tan A \tan B}. $$ On the other hand $$ \tan D=\tan(2\pi -A - B - C)=-\tan(A + B + C) $$ So it is remains to show that $$ \tan(A + B + C)=\frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan B \tan C - \tan A \tan C - \tan A \tan B} $$ which was done here.
$$A+B+C+D=2\pi\iff A+B=2\pi-(C+D)$$
$$\implies\tan(A+B)=\tan\{2\pi-(C+D)\}=-\tan(C+D)$$
Using Addition formula, $$\frac{\tan A+\tan B}{1-\tan A\tan B}=-\frac{\tan C+\tan D}{1-\tan C\tan D}$$
On rearrangement, $$\sum \tan A=\sum \tan A\tan B\tan C=\prod\tan A\left(\sum\frac1{\tan A}\right)$$
$$\sum \tan A=\left(\prod\tan A\right)\left(\sum\cot A\right)$$
Divide either sides by $\prod\tan A$ which is non-zero and finite as no angle is multiple of $\displaystyle\frac\pi2$
Interestingly, $\displaystyle A+B+C+D=n\pi$ (where $n$ is any integer) will satisfy this relation
