If $\cos \frac{x}{2}.\cos \frac{x}{2^2}.\cos \frac{x}{2^3}\mathrm{.....}\infty =\frac{\sin x}{x}$ , then $\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+\mathrm{....}=$

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

$cotx=\frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{2\tan {\displaystyle \frac{x}{2}}}$

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

$cot\frac{x}{2}=\frac{cot{\displaystyle \frac{x}{4}}}{2}-\frac{\tan {\displaystyle \frac{x}{4}}}{2}$

$\frac{1}{2}cot\frac{x}{2}=\frac{cot{\displaystyle \frac{x}{4}}}{4}-\frac{\tan {\displaystyle \frac{x}{4}}}{4}$--------(2)

$similarly \frac{1}{2^{n-1}}cot\frac{x}{2}=\frac{cot{\displaystyle \frac{x}{2^n}}}{2^n}-\frac{\tan {\displaystyle \frac{x}{2^n}}}{2^n}$----------(n)

adding all the equations we get$\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+.....\frac{1}{2^n}\tan \frac{x}{2^n}=\frac{1}{2^n}cot\frac{x}{2^n}-cotx$

$\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+.........\infty =\underset{n\to \infty }{\lim }\left(\frac{1}{2^n}cot\frac{x}{2^n}-cotx\right)$

$=\underset{n\to \infty }{\lim }\left(\frac{1}{2^n}\frac{\cos \frac{x}{2^n}}{\sin \frac{x}{2^n}}-cotx\right)$

$=\underset{n\to \infty }{\lim }\cos \frac{x}{2^n}\frac{1}{2^n}\frac{1}{{\displaystyle \frac{\sin \frac{x}{2^n}}{{\displaystyle \frac{x}{2^n}}}}\frac{x}{2^n}}-cotx$

$=\underset{n\to \infty }{\lim }\cos \frac{x}{2^n} \underset{\frac{x}{2^n}\to 0}{\lim }\frac{1}{\frac{\sin \frac{x}{2^n} }{{\displaystyle \frac{x}{2^n}}}x}-cotx$

$=\frac{1}{x}-$cotx