An equality involving trigonometry I am not able to prove the following equality:
$$\frac{\sin12°\sin48°\sin18°}{\sin84°\sin18°+\sin12°\sin 48°\cos18°}=\tan6°$$
Please help.
Thanks!
 A: We need to prove that
$$2\cos^26^{\circ}\sin48^{\circ}\sin18^{\circ}=\cos6°\sin18°+\sin12°\sin 48°\cos18°$$ or
$$2\cos6^{\circ}\sin48^{\circ}\sin18^{\circ}=\sin18°+2\sin6°\sin 48°\cos18°$$ or
$$2\sin48^{\circ}\sin12^{\circ}=\sin18^{\circ}$$ or
$$\cos36^{\circ}-\frac{1}{2}=\sin18^{\circ}$$ or
$$\cos36^{\circ}-\cos72^{\circ}=\frac{1}{2}$$ or
$$2\cos18^{\circ}\cos36^{\circ}-2\cos18^{\circ}\cos72^{\circ}=\cos18^{\circ}$$ or
$$\cos18^{\circ}+\cos54^{\circ}-\cos90^{\circ}-\cos54^{\circ}=\cos18^{\circ},$$
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:
It is necessary & sufficient to prove 
$$\frac{\sin84°\sin18°+\sin12°\sin 48°\cos18°}{\sin12°\sin48°\sin18°}=\dfrac1{\tan6^\circ}$$
$$\text{Now, }\frac{\sin84°\sin18°}{\sin12°\sin48°\sin18°}=\dfrac{\cos(90-84)^\circ}{2\sin6^\circ\cos6^\circ\sin48^\circ}=\dfrac1{2\sin6^\circ\sin48^\circ}$$
$$\text{and }\frac{\sin12°\sin 48°\cos18°}{\sin12°\sin48°\sin18°}=\cot18^\circ$$
$$\text{Again, }\cot6^\circ-\cot18^\circ=\dfrac{\sin(18-6)^\circ}{\sin6^\circ\sin18^\circ}=\dfrac{2\sin6^\circ\cos6^\circ}{\sin6^\circ\sin18^\circ}=\dfrac{2\cos6^\circ}{\cos72^\circ}$$
So, we need to establish $$\dfrac1{2\sin6^\circ\sin48^\circ}=\dfrac{2\cos6^\circ}{\cos72^\circ}$$ $$\iff\cos72^\circ=2(2\sin6^\circ\cos6^\circ)\sin48^\circ=2\sin12^\circ\sin48^\circ=\cos36^\circ-\cos60^\circ$$
which has been proved here : Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
