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?

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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.

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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.

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