If $3A\ne (2n+1)\frac{\pi }{2}$  prove that  $\tan A\tan \left({60}^{\circ }+A\right)\tan \left({60}^{\circ }-A\right)=\tan 3A$  and find the value of 

$\tan 6^{\circ }\tan {42}^{\circ }\tan {66}^{\circ }\tan {78}^{\circ }$

We know that 

$\mathrm{Sin}A\mathrm{Sin}\left({60}^{\circ }+A\right)\mathrm{Sin}\left({60}^{\circ }-A\right)=\frac{1}{4}\mathrm{Si}n3A$ .......(1)

and 

$\mathrm{Cos}A\mathrm{Cos}\left({60}^{\circ }+A\right)\mathrm{Cos}\left({60}^{\circ }-A\right)=\frac{1}{4}\mathrm{Cos}3A$.....(2)

on dividing (1) and (2) we get 

$\Rightarrow \mathrm{Tan}A\cdot \mathrm{Tan}\left({60}^{\circ }+A\right)\mathrm{Tan}\left({60}^{\circ }-\mathrm{A}\right)=\mathrm{Tan}3\mathrm{A}$ ......(3)

Put $A=6^0$  in (3)

$\Rightarrow \mathrm{Tan}{6}^{\circ }\mathrm{Tan}{66}^{\circ }\mathrm{Tan}{54}^{\circ }=\mathrm{Tan}{18}^{\circ }$.....(4)

Again substitute $A={18}^{\circ }$   in (3)

$\Rightarrow \mathrm{Tan}{18}^{\circ }\mathrm{Tan}{78}^{\circ }\mathrm{Tan}{42}^{\circ }=\mathrm{Tan}54$ ....(5)

Multiply (4) and (5)

$\Rightarrow \mathrm{Tan}{6}^{\circ }\mathrm{Tan}{42}^{\circ }\mathrm{Tan}{66}^{\circ }\mathrm{Tan}{78}^{\circ }=1$