How to prove $\tan3°\tan63°\tan69°=\tan15°$? 
Prove $\tan3°\tan63°\tan69°=\tan15°$


And assuming we don't know that $\tan15^{\circ}$ part, how to just evaluate $\tan3^{\circ} tan63^{\circ} tan69^{\circ}$?
 A: We need to prove that
$$\sin3^{\circ}\sin63^{\circ}\sin69^{\circ}\cos15^{\circ}=\cos3^{\circ}\cos63^{\circ}\cos69^{\circ}\sin15^{\circ}$$ or
$$(\cos60^{\circ}-\cos66^{\circ})(\sin84^{\circ}+\sin54^{\circ})=(\cos60^{\circ}+\cos66^{\circ})(\sin84^{\circ}-\sin54^{\circ})$$ or
$$\sin84^{\circ}\cos66^{\circ}=\sin54\cos60^{\circ}$$ or
$$\sin150^{\circ}+\sin18^{\circ}=\sin54^{\circ}$$ or
$$\frac{1}{2}+\frac{\sqrt5-1}{4}=\frac{\sqrt5+1}{4},$$
which is obvious.
Done!
A: Though I hate reverse engineering, I'm yet to find how the problem came being.
Here is one approach of validating the proposition:
We need $$\dfrac{\sin3^\circ\sin63^\circ}{\cos3^\circ\cos63^\circ}=\dfrac{\sin15^\circ\sin21^\circ}{\cos15^\circ\cos21^\circ}$$
Using componendo and dividendo,  $$\dfrac{\cos(63-3)^\circ}{\cos(63+3)^\circ}=\dfrac{\cos(21-15)^\circ}{\cos(21+15)^\circ}$$
$$\iff\cos36^\circ=2\cos6^\circ\cos66^\circ=\cos(66-6)^\circ+\cos(66+6)^\circ$$
which has been established here: Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
