How would you prove that $\tan({\alpha}) +2\tan({2\alpha})+4\tan({4\alpha})+8\cot (8\alpha) = \cot(\alpha )$? How would you prove that $\tan({\alpha}) +2\tan({2\alpha})+4\tan({4\alpha})+8\cot (8\alpha) = \cot(\alpha )$? 
Regards!

You have:
$$\require{cancel}\cot (2 a)=\frac{\cot (a)}{2}-\frac{\tan (a)}{2}$$
Thus:
$$\begin{align}
\cot a\stackrel{?}{=}&\tan a+2\tan(2a)+4\tan(4a)+8\left(\frac{1}{2} \cot (4 a)-\frac{1}{2} \tan (4 a)\right)\\
\stackrel{?}{=}& \tan a+2\tan (2a)+\cancel{4\tan(4a)}+4\cot (4a)-\cancel{4\tan(4a)}\\
\stackrel{?}{=}&\tan a+2\tan(2a)+4\left(\frac{1}{2} \cot (2 a)-\frac{1}{2} \tan (2 a)\right)\\
\stackrel{?}{=}&\tan a+\cancel{2\tan(2a)}+2\cot(2a)-\cancel{2\tan(2a)}\\
\stackrel{?}{=}&\tan a+2\left(\frac{\cot (a)}{2}-\frac{\tan (a)}{2}\right)\\
\stackrel{?}{=}&\cancel{\tan a}+\cot a-\cancel{\tan a}\\
\cot a\stackrel{\checkmark}{=}&\cot a
\end{align}$$
