Explanation

(c)
We have$,\tan{20{^\circ}\tan{40{^\circ}\tan{60{^\circ}\tan{80{^\circ}} = \frac{\sin{20{^\circ}\sin{40{^\circ}\sin{80{^\circ}}}}}{\cos{20{^\circ}\cos{40{^\circ}\cos{80{^\circ}}}}}\tan{60{^\circ}}}}}$
Here, numerator$\ = (\sin{20{^\circ}\sin{40{^\circ}\sin{80{^\circ})}}}$
$= \frac{\sin{20{^\circ}}}{2}(2\sin{40{^\circ}\sin{80{^\circ})}}$
$= \frac{\sin{20{^\circ}}}{2}(\cos{40{^\circ} - cos120{^\circ})}$
$= \frac{1}{2}\sin{20{^\circ}\left( 1 - 2\sin^{2}{20{^\circ} + \frac{1}{2}} \right)}$
$= \frac{1}{2}\sin{20{^\circ}\left( \frac{3}{2} - 2\sin^{2}{20{^\circ}} \right)}$
$= \frac{1}{4}\left\lbrack 3\sin{20{^\circ} - 4\sin^{3}{20{^\circ}}} \right\rbrack$
$= \frac{\sin{60{^\circ}}}{4} = \frac{\sqrt{3}}{8}$
Now, denominator$\ = \cos{20{^\circ}\cos{40{^\circ}\cos{80{^\circ}}}}$
$= \frac{\sin{2^{3}20{^\circ}}}{2^{3}\sin{20{^\circ}}} = \frac{\sin{160{^\circ}}}{8\sin{20{^\circ}}}$
$= \frac{\sin{20{^\circ}}}{8\sin{20{^\circ}}} = \frac{1}{8}$
Hence,$\ tan20{^\circ}\tan{40{^\circ}\tan{80{^\circ} = \frac{\frac{\sqrt{3}}{8}}{\frac{1}{8}} = \sqrt{3}}}$
$\Rightarrow \tan{20{^\circ}\tan{40{^\circ}\tan{60{^\circ}\tan{80{^\circ} = \sqrt{3}} \bullet \sqrt{3}}}} = 3$