trigonometry - 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?

Answer

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.