My favourite trigonometric identity

I enjoyed proving trigonometric identities when I was in high school. The following identity that appeared on an old mathematics contest (Descartes contest) is perhaps my favourite one as it has a neat formulation:

For triangle $ABC$ , we have:

$\tan A + \tan B + \tan C = \tan A \tan B \tan C .$

Here is the proof:

$\begin{align*} & \tan A + \tan B + \tan C \\ = & \big( \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} \big) + \frac{\sin C}{\cos C} \\ = & \big( \frac{\sin A\cos B + \sin B\cos A}{\cos A\cos B} \big) +\frac{\sin C}{\cos C} \\ = & \frac{\sin (A+B)}{\cos A\cos B} + \frac{\sin C}{\cos C}.\\ \end{align*}$

Since $A+B=\pi-C$ , we have $\sin (A+B)=\sin C$ . Hence

$\begin{align*} & \tan A + \tan B + \tan C \\ = & \frac{\sin C}{\cos A\cos B} + \frac{\sin C}{\cos C} \\ = &\frac{\sin C}{\cos A\cos B\cos C}(\cos C + \cos A\cos B).\\ \end{align*}$

Since $A+B=\pi-C$ , we have $\cos (C)=-\cos(A+B).$ Thus

$\begin{align*} & \tan A + \tan B + \tan C \\ = &\frac{\sin C}{\cos A\cos B\cos C}(-\cos(A+B) + \cos A\cos B) \\ =&\frac{\sin A\sin B\sin C}{\cos A\cos B\cos C} \\ =&\tan A\tan B\tan C. \\ \end{align*}$