tan 20^circ tan 40^circ tan 60^circ tan 80^ci | vedclass

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Question

(c) $	an \,\,{20^o}	an \,\,{40^o}	an \,\,{60^o}	an \,\,{80^o}$

$ = \frac{{\sin \,\,{{20}^o}\sin \,\,{{40}^o}\sin \,\,{{80}^o}	an {{60}^o}}}{{\

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(c) $	an \,\,{20^o}	an \,\,{40^o}	an \,\,{60^o}	an \,\,{80^o}$

$ = \frac{{\sin \,\,{{20}^o}\sin \,\,{{40}^o}\sin \,\,{{80}^o}	an {{60}^o}}}{{\cos \,\,{{20}^o}\cos \,\,{{40}^o}\cos \,\,{{80}^o}}}$

Here ${N^r} = (\sin \,\,{20^o}\sin \,\,{40^o}\sin \,\,{80^o})$

$ = \frac{{\sin \,\,{{20}^o}}}{2}\,(2\,\,\sin \,\,{40^o}\sin \,\,{80^o})$

$ = \frac{{\sin \,\,{{20}^o}}}{2}\,(\cos \,\,{40^o} - \cos \,\,{120^o})$

$ = \frac{1}{2}\sin \,\,{20^o}\,\left( {1 - 2\,\,{{\sin }^2}{{20}^o} + \frac{1}{2}} ight)$

$ = \frac{1}{2}\sin \,\,{20^o}\,\left( {\frac{3}{2} - 2\,\,{{\sin }^2}{{20}^o}} ight) = \frac{{\sin \,{{60}^o}}}{4} = \frac{{\sqrt 3 }}{8}$

Now, we take ${D^r} = \cos {20^o}\cos {40^o}\cos {80^o}$

$ = \frac{{\sin \,\,{2^3}\,{{20}^o}}}{{{2^3}\,\sin \,\,{{20}^o}}} = \frac{{\sin \,\,{{160}^o}}}{{8\,\,\sin \,\,{{20}^o}}} = \frac{{\sin \,\,{{20}^o}}}{{8\,\,\sin \,\,{{20}^o}}} = \frac{1}{8}$

$	herefore $ Hence $	an \,\,{20^o}	an \,\,{40^o}	an \,\,{80^o} = \frac{{\sqrt 3 /8}}{{1/8}}$

Therefore $	an {20^o}	an {40^o}	an {60^o}	an {80^o} = \sqrt 3 .\sqrt 3 = 3$.

$\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = $

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