TITLE: Prove $\tan 84^{\circ}=\tan 78^{\circ}+\tan 72^{\circ}+\tan 60^{\circ}$
QUESTION [5 upvotes]: Prove that: $$\tan 84^{\circ}=\tan 78^{\circ}+\tan 72^{\circ}+\tan 60^{\circ}$$
First I simplified $\tan 78^{\circ}+\tan 72^{\circ}$ and then I simplified $\tan 84^{\circ}-\tan 60^{\circ}$. 
Both turned out to be $$\frac{1}{2\cos 72^{\circ}\cos 78^{\circ}}$$
Just usual formulae bashing.
Is there a more elegant way.

REPLY [4 votes]: Note that
$$\cos36-\cos72 = 2\cos36\cos 72= \frac{\sin72\cos72}{\sin36}=\frac{\sin144}{2\sin36}=\frac12$$
Thus
\begin{align}
&\tan 78+\tan 72+\tan 60\\
=&\frac{1}{2\cos72\cos78}+ \cot30=\frac{2\cos36}{\cos78} +2{\cos30}\\
=& \frac{2\cos36+2\cos30\cos78}{\cos78}
=\frac{2\cos36-\cos72+\cos48}{\sin12}\\
=&\frac{1+\cos72+\cos48}{\sin12} =\frac{1+\cos12}{\sin12}\\
=&\frac{2\cos^26}{2\sin6\cos6}=\tan84
\end{align}