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.