Let $f\left(x\right)=\frac{2x(\sin x+\tan x)}{2\left[\frac{x+21\pi }{\pi }\right]-41},x\ne n\pi$, then f is (where [ . ] represents greatest integer function)

The denominator is

$=2\left[\frac{x+21\pi }{\pi }\right]-41=2\left[\frac{x}{\pi }+21\right]-41$

$=2\left\{21+\left[\frac{x}{\pi }\right]\right\}-41=2\left[\frac{x}{\pi }\right]+1$

$\therefore f\left(x\right)=\frac{x(\sin x+\tan x)}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}$

$\Rightarrow f(-x)=\frac{-x\left\{\sin (-x)+\tan (-x)\right\}}{\left[-\frac{x}{\pi }\right]+\frac{1}{2}}=\frac{x(\sin x+\tan x)}{-1-\left[\frac{x}{\pi }\right]+\frac{1}{2}}$

$=-\frac{x(\sin x+\tan x)}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}=-f\left(x\right)$