Find the value of $$9\tan{10^{\circ}}+2\tan{20^{\circ}}+4\tan{40^{\circ}}-\tan{80^{\circ}}$$
It seem the answer is $0$, see http://www.wolframalpha.com/input/?i=9tan(10)%2B2tan(20)%2B4tan(40)-tan(80).
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math110
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See https://math.stackexchange.com/questions/124528/showing-the-identity-tan-alpha-2-tan-2-alpha-4-tan-4-alpha-cot-alp and https://math.stackexchange.com/questions/422087/solve-tan%ce%b12-tan2%ce%b14-tan4%ce%b18-tan8%ce%b116-tan%ce%b1-cot%ce%b1-for-alpha?noredirect=1&lq=1 – lab bhattacharjee Dec 12 '17 at 15:44
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with the help of $$\tan \theta = \cot \theta-2\cot 2\theta$$
Replace $\theta \rightarrow 2\theta$ , we have $$2\tan 2\theta =2\cot 2\theta -4\cot(4\theta)$$
similarly $$4\tan(4\theta) = 4\cot(4\theta)-4\cot(8\theta)$$
So $$\tan \theta+2\tan 2\theta+2\tan 4\theta = -8\cot(8\theta)$$
put $\theta = 10^\circ$
$$\tan 10^\circ+2\tan 20^\circ+4\tan 40^\circ =\cot (10^\circ) -8\cot(80^\circ) = \tan(80^\circ)-8\tan(10^\circ)$$
$$9\tan 10^\circ+2\tan 20^\circ+4\tan 40^\circ -\tan (10^\circ) = 0$$
DXT
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$\tan\theta + 2\tan(2\theta) + 2\tan(4\theta) \neq -8\cot(8\theta)$. Did you mean to write $\tan\theta + 2\tan(2\theta) + 4\tan(4\theta) = \cot\theta - 8\cot(8\theta)$? – N. F. Taussig Dec 12 '17 at 13:01