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.
$\endgroup$1 Answer
$\begingroup$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}
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