I think proving a more general form will actually be easier than this specific case - it appears to be true that
[tex]2^{n+1}\cot(2^{n+1}\alpha)+\displaystyle\sum_{i=0}^n2^i\tan(2^i\alpha)=\cot\alpha[/tex]
for [tex]n=0,1,2,3,\ldots[/tex].
Let's consider a proof by induction. The base case [tex]n=0[/tex] gives
[tex]2\cot2\alpha+\tan\alpha=2\dfrac{\cos2\alpha}{\sin2\alpha}+\dfrac{\sin\alpha}{\cos\alpha}[/tex]
[tex]=\dfrac{\cos^2\alpha-\sin^2\alpha}{\sin\alpha\cos\alpha}+\dfrac{\sin\alpha}{\cos\alpha}[/tex]
[tex]=\dfrac{\cos^2\alpha-\sin^2\alpha+\sin^2\alpha}{\sin\alpha\cos\alpha}[/tex]
[tex]=\dfrac{\cos\alpha}{\sin\alpha}=\cot\alpha[/tex]
as desired.
Suppose the identity holds for [tex]n=k[/tex], so that
[tex]2^{k+1}\cot(2^{k+1}\alpha)+\displaystyle\sum_{i=0}^k2^i\tan(2^i\alpha)=\cot\alpha[/tex]
For [tex]n=k+1[/tex], we have
[tex]2^{k+2}\cot(2^{k+2}\alpha)+\displaystyle\sum_{i=0}^{k+1}2^i\tan(2^i\alpha)[/tex]
[tex]=2^{k+2}\cot(2^{k+2}\alpha)+2^{k+1}\tan(2^{k+1}\alpha)+\displaystyle\sum_{i=0}^k2^i\tan(2^i\alpha)[/tex]
[tex]=2^{k+2}\cot(2^{k+2}\alpha)+2^{k+1}\tan(2^{k+1}\alpha)+(\cot\alpha-2^{k+1}\cot(2^{k+1}\alpha))[/tex]
So we ultimately need to show that
[tex]2^{k+2}\cot(2^{k+2}\alpha)+2^{k+1}\tan(2^{k+1}\alpha)-2^{k+1}\cot(2^{k+1}\alpha)=0[/tex]
or
[tex]2\cot(2^{k+2}\alpha)+\tan(2^{k+1}\alpha)=\cot(2^{k+1}\alpha)[/tex]
If we replace [tex]\beta=2^{k+1}\alpha[/tex], we get(!) the base case, which we've shown to be true,
[tex]2\cot2\beta+\tan\beta=\cot\beta[/tex]
and thus the identity is proved.
