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Question: \[\tan 20{^\circ}\tan 40{^\circ}\tan 60{^\circ}\tan 80{^\circ} =\]...

tan⁡20∘tan⁡40∘tan⁡60∘tan⁡80∘=\tan 20{^\circ}\tan 40{^\circ}\tan 60{^\circ}\tan 80{^\circ} =

A

1

B

2

C

3

D

3/2\sqrt{3}/2

Answer

3

Explanation

Solution

tan⁡20otan⁡40otan⁡60otan⁡80o\tan{}20^{o}\tan{}40^{o}\tan{}60^{o}\tan{}80^{o}

=sin⁡20osin⁡40osin⁡80otan⁡60ocos⁡20ocos⁡40ocos⁡80o= \frac{\sin 20^{o}\sin 40^{o}\sin 80^{o}\tan 60^{o}}{\cos 20^{o}\cos 40^{o}\cos 80^{o}}

Here Nr=(sin⁡20osin⁡40osin⁡80o)N^{r} = (\sin{}20^{o}\sin{}40^{o}\sin{}80^{o})

=sin⁡20o2(2sin⁡40osin⁡80o)= \frac{\sin 20^{o}}{2}(2\sin{}40^{o}\sin{}80^{o})

=sin⁡20o2(cos⁡40o−cos⁡120o)= \frac{\sin 20^{o}}{2}(\cos{}40^{o} - \cos{}120^{o})

=12sin⁡20o(1−2sin⁡220o+12)= \frac{1}{2}\sin{}20^{o}\left( 1 - 2\sin^{2}20^{o} + \frac{1}{2} \right)

=12sin⁡20o(32−2sin⁡220o)=sin⁡60o4=38= \frac{1}{2}\sin{}20^{o}\left( \frac{3}{2} - 2\sin^{2}20^{o} \right) = \frac{\sin 60^{o}}{4} = \frac{\sqrt{3}}{8}

Now, we take Dr=cos⁡20ocos⁡40ocos⁡80oD^{r} = \cos 20^{o}\cos 40^{o}\cos 80^{o}

=sin⁡2320o23sin⁡20o=sin⁡160o8sin⁡20o=sin⁡20o8sin⁡20o=18= \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}

∴\therefore Hence tan⁡20otan⁡40otan⁡80o=3/81/8\tan{}20^{o}\tan{}40^{o}\tan{}80^{o} = \frac{\sqrt{3}/8}{1/8}

Thereforetan⁡20otan⁡40otan⁡60otan⁡80o=3.3=3\tan 20^{o}\tan 40^{o}\tan 60^{o}\tan 80^{o} = \sqrt{3}.\sqrt{3} = 3