We have to prove that $$ \tan 20°\cdot \tan 40° \cdot \tan 60° \cdot \tan 80° = 3$$
I tried to use the formula
$$\displaystyle\frac {\tan a+\tan b}{1-\tan a\cdot \tan b}$$
But in that I got stuck .
Prove that $ \tan 20°\cdot \tan 40° \cdot \tan 60° \cdot \tan 80° = 3$
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trigonometry
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1Closely related: [How can I find the following product? $ \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.$](http://math.stackexchange.com/questions/805821/how-can-i-find-the-following-product-tan-20-circ-cdot-tan-40-circ-cdot). – 2017-01-25
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0See http://math.stackexchange.com/questions/455070/proving-a-fact-tan6-circ-tan42-circ-tan12-circ-tan24-cir – 2017-01-25
1 Answers
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Hint:
$$\tan\alpha\tan(60^{\circ}-\alpha)\tan(60^{\circ}+\alpha)=\tan3\alpha$$ If $\alpha=20^{\circ}$, then $$\tan20^{\circ}\tan40^{\circ}\tan80^{\circ}=\tan60^{\circ}=\sqrt3$$
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0yeah , correct hint – 2017-01-25