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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!

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2 Answers 2

up vote 4 down vote accepted

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.

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Brilliant! $\mathrm{}$ –  Ahaan S. Rungta Nov 27 '13 at 23:31

$$A+B+C+D=2\pi\iff A+B=2\pi-(C+D)$$

$$\implies\tan(A+B)=\tan\{2\pi-(C+D)\}=-\tan(C+D)$$

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

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@AhaanRungta, how about this one? –  lab bhattacharjee Nov 28 '13 at 4:34
    
Also very nice! Unfortunately, I cannot accept more than one answer. :( –  Ahaan S. Rungta Nov 28 '13 at 15:00
    
@AhaanRungta, acceptance is your call. I just wanted share an answer which needs lesser formula & calculations. –  lab bhattacharjee Nov 28 '13 at 15:03
    
It is indeed nice, which is why I gave the +1. :) –  Ahaan S. Rungta Nov 28 '13 at 15:07
    
@AhaanRungta, thanks for your feedback & also for the question –  lab bhattacharjee Nov 28 '13 at 15:08

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