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!
 A: You have:
$$\require{cancel}\cot (2 a)=\frac{\cot (a)}{2}-\frac{\tan (a)}{2}$$
Thus:
$$\begin{align}
\cot a\stackrel{?}{=}&\tan a+2\tan(2a)+4\tan(4a)+8\left(\frac{1}{2} \cot (4 a)-\frac{1}{2} \tan (4 a)\right)\\
\stackrel{?}{=}& \tan a+2\tan (2a)+\cancel{4\tan(4a)}+4\cot (4a)-\cancel{4\tan(4a)}\\
\stackrel{?}{=}&\tan a+2\tan(2a)+4\left(\frac{1}{2} \cot (2 a)-\frac{1}{2} \tan (2 a)\right)\\
\stackrel{?}{=}&\tan a+\cancel{2\tan(2a)}+2\cot(2a)-\cancel{2\tan(2a)}\\
\stackrel{?}{=}&\tan a+2\left(\frac{\cot (a)}{2}-\frac{\tan (a)}{2}\right)\\
\stackrel{?}{=}&\cancel{\tan a}+\cot a-\cancel{\tan a}\\
\cot a\stackrel{\checkmark}{=}&\cot a
\end{align}$$
A: After a little simplification,
$$t+2\frac{2t}{1-t^2}+4\frac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}+8\dfrac{1-\left(\dfrac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}\right)^2}{2\dfrac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}}
=\frac1t.$$
A: 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
$$\tan({\alpha}) +2\tan({2\alpha})+4\tan({4\alpha})+8\cot (8\alpha) = \cot(\alpha )$$
