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

Question

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

Regards!

Answer 1

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

Answer 2

After a little simplification,

$$t+2\frac{2t}{1-t^2}+4\frac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}+8\dfrac{1-\left(\dfrac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}\right)^2}{2\dfrac{2\dfrac{2t}{1-t^2}}{1-\left(\dfrac{2t}{1-t^2}\right)^2}} =\frac1t.$$

Answer 3

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 $$\tan({\alpha}) +2\tan({2\alpha})+4\tan({4\alpha})+8\cot (8\alpha) = \cot(\alpha )$$