Trigonometric Identities Question 52
Question: $ \tan 20{}^\circ \tan 40{}^\circ \tan 60{}^\circ \tan 80{}^\circ = $
[IIT 1974]
Options:
A) 1
B) 2
C) 3
D) $ \sqrt{3}/2 $
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Answer:
Correct Answer: C
Solution:
$ \tan 20^{o}\tan 40^{o}\tan 60^{o}\tan 80^{o} $ $ =\frac{\sin 20^{o}\sin 40^{o}\sin 80^{o}\tan 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},( 1-2{{\sin }^{2}}20^{o}+\frac{1}{2} ) $ $ =\frac{1}{2}\sin 20^{o},( \frac{3}{2}-2{{\sin }^{2}}20^{o} )=\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} $
$ \therefore $ Hence $ \tan 20^{o}\tan 40^{o}\tan 80^{o}=\frac{\sqrt{3}/8}{1/8} $ Therefore $ \tan 20^{o}\tan 40^{o}\tan 60^{o}\tan 80^{o}=\sqrt{3}.\sqrt{3}=3 $ .
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