How would you prove that $\tan({\alpha}) +2\tan({2\alpha})+4\tan({4\alpha})+8\cot (8\alpha) = \cot(\alpha )$?
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Could you give more contexte ? as : is it really an exercise or you found this formula somewhere in a book or somewhere else ? Because if you found it somewhere, there is probably not an other way than computing the LHS and RHS brute force and conclude. But if an exercise ask you to prove that formula, there is probably a trick... But anyway, I tried brute force, and it's not that huge to do it (just use the fact that $\tan(2\alpha )=\frac{2\tan(\alpha )}{1-\tan^2(\alpha )}$), but it's probably not the aim of the exercise... – Surb Jul 10 '18 at 15:49
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You have: $$\require{cancel}\cot (2 a)=\frac{\cot (a)}{2}-\frac{\tan (a)}{2}$$ Thus: $$\begin{align} \cot a\stackrel{?}{=}&\tan a+2\tan(2a)+4\tan(4a)+8\left(\frac{1}{2} \cot (4 a)-\frac{1}{2} \tan (4 a)\right)\\ \stackrel{?}{=}& \tan a+2\tan (2a)+\cancel{4\tan(4a)}+4\cot (4a)-\cancel{4\tan(4a)}\\ \stackrel{?}{=}&\tan a+2\tan(2a)+4\left(\frac{1}{2} \cot (2 a)-\frac{1}{2} \tan (2 a)\right)\\ \stackrel{?}{=}&\tan a+\cancel{2\tan(2a)}+2\cot(2a)-\cancel{2\tan(2a)}\\ \stackrel{?}{=}&\tan a+2\left(\frac{\cot (a)}{2}-\frac{\tan (a)}{2}\right)\\ \stackrel{?}{=}&\cancel{\tan a}+\cot a-\cancel{\tan a}\\ \cot a\stackrel{\checkmark}{=}&\cot a \end{align}$$
John Glenn
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After a little simplification,
$$t+2\frac{2t}{1-t^2}+4\frac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}+8\dfrac{1-\left(\dfrac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}\right)^2}{2\dfrac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}} =\frac1t.$$
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Simply note that $$\cot\alpha-2\cot2\alpha=\tan\alpha \tag{1}$$ $$2\cot2\alpha-4\cot4\alpha=2\tan2\alpha \tag{2}$$ $$4\cot2\alpha-8\cot4\alpha=4\tan8\alpha \tag{3}$$ By adding $(1)+(2)+(3)$, we get $$\tan({\alpha}) +2\tan({2\alpha})+4\tan({4\alpha})+8\cot (8\alpha) = \cot(\alpha )$$
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Hope you don't mind, I cleaned up your LaTex a bit by adding "tags". Feel free to change it back if you don't think it's an improvement. – awkward Jul 10 '18 at 18:21