<s>[INST] Prove by induction: $\sum\limits_{k=1}^{n}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{2^n}\cot\frac{x}{2^n}-\cot x,x\neq k\pi,k\in \mathbb{Z}$ $\sum\limits_{k=1}^{n}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{2^n}\cot\frac{x}{2^n}-\cot x,x\neq k\pi,k\in \mathbb{Z}$
Base Case: For $n=1$, $\frac{1}{2}\tan\frac{x}{2}=\frac{1}{2}\cot\frac{x}{2}-\cot x$
Proof:
$$\frac{1}{2}\cot\frac{x}{2}-\cot x=\frac{1}{2\tan\frac{x}{2}}-\frac{1}{\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}}=\frac{1-1+\tan^2\frac{x}{2}}{2\tan\frac{x}{2}}=\frac{1}{2}\tan\frac{x}{2}$$
Induction Hypothesis:
Assume true that for $n = m$,
$$\sum\limits_{k=1}^{m}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{2^m}\cot\frac{x}{2^m}-\cot x$$ 
Inductive Step: For $n=m+1$
$\sum\limits_{k=1}^{m+1}\frac{1}{2^{k}}\tan\frac{x}{2^{k}}=\frac{1}{2^{m+1}}\cot\frac{x}{2^{m+1}}-\cot x$
Proof:
$$\frac{1}{2^m}\cot\frac{x}{2^m}-\cot x+\frac{1}{2^{m+1}}\tan\frac{x}{2^{m+1}}=\frac{1}{2^m}\cot\frac{x}{2^m}-\cot x+\frac{1}{2^{m+1}\cot\frac{x}{2^{m+1}}}$$
I don't know how to transform $\frac{1}{2^m}\cot\frac{x}{2^m}-\cot x+\frac{1}{2^{m+1}\cot\frac{x}{2^{m+1}}}$ into $\frac{1}{2^{m+1}}\cot\frac{x}{2^{m+1}}-\cot x$
 [/INST] Proving that $$\frac{1}{2^k}\cot\frac{x}{2^k}-\cot x+\frac{1}{2^{k+1}}\tan\frac{x}{2^{k+1}}=\frac{1}{2^{k+1}}\cot\frac{x}{2^{k+1}}-\cot x$$ is equivalent to prove that $$\frac{1}{2^k}\cot\frac{x}{2^k}+\frac{1}{2^{k+1}}\tan\frac{x}{2^{k+1}}=\frac{1}{2^{k+1}}\cot\frac{x}{2^{k+1}}$$ by omitting $-\cot x$. Then if you multiply both sides by $2^{k+1}$ we get $$2\cot\frac{x}{2^k}+\tan\frac{x}{2^{k+1}}=\cot\frac{x}{2^{k+1}}$$
You proved earlier (case n=1) that $$\tan\frac{y}{2}=\cot\frac{y}{2}-2\cot y$$ Thus if you take $y=\frac{x}{2^k}$ you get $$\tan\frac{x}{2^{k+1}}=\cot\frac{x}{2^{k+1}}-2\cot\frac{x}{2^k}$$
Remark: $k$ is already used for indexation in $\sum$ so you should take a different variable like $i$ or even simply work with $n$ and then you take $n+1$ in the inductive step.
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