Prove this trigonometric identity in quadrilateral If $\alpha,\beta,\gamma,\delta$ are angles in quadrilateral different from $90^\circ$, prove the following: 
$$ \frac{\tan\alpha+\tan\beta+\tan\gamma+\tan\delta}{\tan\alpha\tan\beta\tan\gamma\tan\delta}=\cot\alpha+\cot\beta+\cot\gamma+\cot\delta $$
I tried different transformations with using $\alpha+\beta+\gamma+\delta=2\pi$ in equation above, but no success. Am I missing some not-so-well-known formula? 
 A: It follows directly from $\tan(\alpha + \beta + \gamma + \delta) = 0$ and the sum angle formula for $\tan$ (see here: Tangent sum using symmetric polynomials) 
Using that formula we get (from numerator = 0) that
$$ \tan \alpha + \tan \beta + \tan \gamma + \tan \delta = $$
$$\tan \alpha\tan \beta\tan \gamma+ \tan \alpha\tan \beta\tan \delta + \tan \alpha\tan \gamma\tan \delta + \tan \beta\tan \gamma\tan \delta$$
divididing by $ \tan \alpha\tan \beta\tan \gamma\tan \delta$ gives the result.
A: For $\alpha,\beta,\gamma,\delta$ arbitrary, we have $\tan(\alpha+\beta+\gamma+\delta)$
$$
= \frac{\left(\begin{array} {} \tan\alpha + \tan\beta+\tan\gamma + \tan\delta \\
{} - \tan\alpha\tan\beta\tan\gamma - \tan\alpha\tan\beta\tan\delta-\tan\alpha\tan\gamma\tan\delta-\tan\beta\tan\gamma\tan\delta\end{array}\right)}{\left(\begin{array}  {} 1-\tan\alpha\tan\beta -\tan\alpha\tan\delta-\tan\alpha\tan\gamma \\ {}-\tan\beta\tan\gamma-\tan\beta\tan\delta-\tan\gamma\tan\delta \\ {} + \tan\alpha\tan\beta\tan\gamma\tan\delta \end{array}\right)}
$$
In a quadrilateral, we have $\alpha+\beta+\gamma+\delta = 2\pi\text{ radians} = 360^\circ$.  Therefore $\tan(\alpha+\beta+\gamma+\delta)=0$.  Consequently, the numerator must be $0$.  Hence
$$
\begin{align}
& \tan\alpha + \tan\beta+\tan\gamma + \tan\delta \\  \\
& = \tan\alpha\tan\beta\tan\gamma + \tan\alpha\tan\beta\tan\delta +\tan\alpha\tan\gamma\tan\delta+\tan\beta\tan\gamma\tan\delta.
\end{align}
$$
Divide both sides by the product of four tangents and you have the result.
