Question

$\frac{1}{2}\tan \left(\frac{x}{2}\right)+\frac{1}{4}\tan \left(\frac{x}{4}\right)+...+\frac{1}{2^n}\tan \left(\frac{x}{2^n}\right)=\frac{1}{2^n}\cot \left(\frac{x}{2^n}\right)-\cot x$ for all n ∈ N and $0<x<\frac{\mathrm{\pi }}{2}$

Answer 1

We need to prove $\frac{1}{2}\tan \left(\frac{x}{2}\right)+\frac{1}{4}\tan \left(\frac{x}{4}\right)+...+\frac{1}{2^n}\tan \left(\frac{x}{2^n}\right)=\frac{1}{2^n}\cot \left(\frac{x}{2^n}\right)-\cot x$ for all n ∈ N and $0<x<\frac{\mathrm{\pi }}{2}$ using mathematical induction.

For n = 1,

LHS = $\frac{1}{2}\tan \frac{x}{2}$

and

$$\begin{gathered} \mathrm{RHS}=\frac{1}{2}\cot \frac{x}{2}-\cot x=\frac{1}{2\tan {\displaystyle \frac{x}{2}}}-\frac{1}{\tan x} \\ \Rightarrow \mathrm{RHS}=\frac{1}{2\tan {\displaystyle \frac{x}{2}}}-\frac{1}{{\displaystyle \frac{2\tan {\displaystyle \frac{x}{\mathit{2}}}}{1-{\tan }^2{\displaystyle \frac{x}{\mathit{2}}}}}} \\ \Rightarrow \mathrm{RHS}=\frac{1}{2\tan {\displaystyle \frac{x}{2}}}-\frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{{\displaystyle 2\tan \frac{x}{2}}}=\frac{1-1+{\tan }^2{\displaystyle \frac{x}{2}}}{{\displaystyle 2\tan \frac{x}{2}}}=\frac{\tan {\displaystyle \frac{x}{2}}}{2} \end{gathered}$$

Therefore, the given relation is true for n = 1.

Now, let the given relation be true for n = k.

We need to prove that the given relation is true for n = k + 1.

$\therefore \frac{1}{2}\tan \left(\frac{x}{2}\right)+\frac{1}{4}\tan \left(\frac{x}{4}\right)+...+\frac{1}{2^k}\tan \left(\frac{x}{2^k}\right)=\frac{1}{2^k}\cot \left(\frac{x}{2^k}\right)-\cot x$

Now,

$\frac{1}{2}\tan \left(\frac{x}{2}\right)+\frac{1}{4}\tan \left(\frac{x}{4}\right)+...+\frac{1}{2^k}\tan \left(\frac{x}{2^k}\right)+\frac{1}{2^{k+1}}\tan \left(\frac{x}{2^{k+1}}\right)=\frac{1}{2^k}\cot \left(\frac{x}{2^k}\right)-\cot x+\frac{1}{2^{k+1}}\tan \left(\frac{x}{2^{k+1}}\right)$

Let:

$L=\frac{1}{2^k}\cot \left(\frac{x}{2^k}\right)-\cot x+\frac{1}{2^{k+1}}\tan \left(\frac{x}{2^{k+1}}\right)$.

$$\begin{gathered} \Rightarrow L=\frac{1}{2^k}\cot \left(\frac{x}{2^k}\right)+\frac{1}{2^{k+1}}\tan \left(\frac{x}{2^{k+1}}\right)-\cot x \\ \Rightarrow L=\frac{1}{2^k\tan {\displaystyle \left(\frac{x}{2^k}\right)}}+\frac{1}{2^{k+1}}\tan \left(\frac{x}{2^{k+1}}\right)-\cot x \\ \Rightarrow L=\frac{1}{2^k\tan 2{\displaystyle \left(\frac{x}{2^{k+1}}\right)}}+\frac{1}{2^{k+1}}\tan \left(\frac{x}{2^{k+1}}\right)-\cot x \\ \Rightarrow L=\frac{1}{{\displaystyle 2^k\times \frac{{\displaystyle 2\tan \left(\frac{x}{2^{k+1}}\right)}}{1-{\tan }^2\left({\displaystyle \frac{x}{2^{k+1}}}\right){\displaystyle }}}}+\frac{1}{2^{k+1}}\tan \left(\frac{x}{2^{k+1}}\right)-\cot x \\ \Rightarrow L=\frac{1-{\tan }^2{\displaystyle \left(\frac{x}{2^{k+1}}\right)}}{{\displaystyle 2^{k+1}\tan \left(\frac{x}{2^{k+1}}\right)}}+\frac{1}{2^{k+1}}\tan \left(\frac{x}{2^{k+1}}\right)-\cot x \end{gathered}$$

$\Rightarrow L=\frac{1-{\tan }^2{\displaystyle \left(\frac{x}{2^{k+1}}\right)+{\tan }^2\left(\frac{x}{2^{k+1}}\right)}}{{\displaystyle 2^{k+1}\tan \left(\frac{x}{2^{k+1}}\right)}}-\cot x=\frac{1}{{\displaystyle 2^{k+1}}}\cot \left(\frac{x}{2^{k+1}}\right)-\cot x$
Now,

$\frac{1}{2}\tan \left(\frac{x}{2}\right)+\frac{1}{4}\tan \left(\frac{x}{4}\right)+...+\frac{1}{2^k}\tan \left(\frac{x}{2^k}\right)+\frac{1}{2^{k+1}}\tan \left(\frac{x}{2^{k+1}}\right)=\frac{1}{2^{k+1}}\cot \left(\frac{x}{2^{k+1}}\right)-\cot x$

Thus, $\frac{1}{2}\tan \left(\frac{x}{2}\right)+\frac{1}{4}\tan \left(\frac{x}{4}\right)+...+\frac{1}{2^n}\tan \left(\frac{x}{2^n}\right)=\frac{1}{2^n}\cot \left(\frac{x}{2^n}\right)-\cot x$ for all n ∈ N and $0<x<\frac{\mathrm{\pi }}{2}$.