Question

Prove that $\frac{1}{2}tan\left(\frac{x}{2}\right)+\frac{1}{4}tan\left(\frac{x}{4}\right)+...+\frac{1}{2^n}tan\left(\frac{x}{2^n}\right)=\frac{1}{2^n}cot+\left(\frac{x}{2^n}\right)-cotx$ for all $nϵN$ and $0<x<\frac{x}{2}$.

Answer 1

Let P(n) : $\frac{1}{2}tan\left(\frac{x}{2}\right)+\frac{1}{4}tan\left(\frac{x}{4}\right)+...+\frac{1}{2^n}tan\left(\frac{x}{2^n}\right)=\frac{1}{2^n}cot+\left(\frac{x}{2^n}\right)-cotx$

For n = 1

$\frac{1}{2}tan\frac{x}{2}=\frac{1}{2}cot\left(\frac{x}{2}\right)-cotx$

$=\frac{1}{2}\frac{1}{tan\frac{x}{2}}-\frac{1}{tanx}$

$=\frac{1}{2tan\frac{x}{2}}-\frac{1}{(-\frac{2tan\frac{x}{2}}{2-ta{n}^2\frac{x}{2}})}$

$=\frac{1}{2tan\frac{x}{2}}-\frac{1-ta{n}^2\frac{x}{2}}{1-tan\frac{x}{2}}$

$=\frac{1-1+ta{n}^2\frac{x}{2}}{2tan\frac{x}{2}}$

$=\frac{ta{n}^2\frac{x}{2}}{2tan\frac{x}{2}}$

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

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

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

We have to show that,

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

Now,

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

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

$=\frac{1}{2^k}cot\left(\frac{x}{2^k}\right)-cotx+\frac{1}{{2.2}^k}-\frac{1}{cot(\frac{x}{2^k}.\frac{1}{2})}-cotx$

$=\frac{1}{2k}\left(\frac{1-ta{n}^2\left(\frac{x}{2^{k+1}}\right)}{2tan(\frac{x}{2^{k+1}}+\frac{1}{2}tan\left\{\left(\frac{x}{{2.2}^k}\right)\right\})}\right)-cotx$

$=\frac{1}{2^k}(1-ta{n}^2\left(\frac{x}{2^{k+1}}\right)+ta{n}^2\left(\frac{x}{2^{k+1}}\right))-cotx=\frac{1}{2^{k+1}}\left(\frac{1}{tan\left(\frac{x}{2^{k+1}}\right)}\right)-cotx=\frac{1}{2^{k+1}}cot\left(\frac{x}{2^{k+1}}\right)-cotx$

$\Rightarrow P\left(n\right)$ is true for n = k + 1

$\Rightarrow P\left(n\right)$ is true for all $nϵN$ by PMI.