Range of the function $f\left(x\right)=\frac{se{c}^{2}x-tanx}{se{c}^{2}x+tanx}-\frac{\pi }{2}<x<\frac{\pi }{2}$, is

The correct option is C $[\frac{1}{3},3]$
Let

$f\left(x\right)=y$

$y=\frac{se{c}^{2}x-\tan x}{se{c}^{2}x+\tan x}$

$\Rightarrow y(se{c}^{2}x+\tan x)=se{c}^{2}x-\tan x$

$\Rightarrow y(1+{\tan }^{2}x+\tan x)=1+{\tan }^{2}x-\tan x$

$\Rightarrow y+y{\tan }^{2}x+y\tan x=1{\tan }^{2}x-\tan x$

$\Rightarrow {\tan }^{2}x(y-1)+\tan (y+1)+y-1=0$

Now it is given $\frac{-\pi }{2}<x<\frac{\pi }{2}$

So $\tan x$ is real in $x\in (\frac{-\pi }{2},\frac{\pi }{2})$

So $D\ge 0$

$\Rightarrow (y+1{)}^2-4(y-1\left)\right(y-1)\ge 0$

$\Rightarrow y^2+1+2y-4(y-1{)}^2\ge 0$

$\Rightarrow y^2+2y+1-4(y^2+1-2y)\ge 0$

$\Rightarrow y^2+2y+1-4y^2-4+8y\ge 0$

$\Rightarrow -3y^2+10y-3\ge 0$

$\Rightarrow 3y^2-910y+3\le 0$

$\Rightarrow 3y^2-9y-y+3\le 0$

$\Rightarrow 3y(y-3)-1(y-3)\le 0$

$\Rightarrow (3y-1)(y-3)\le 0$