If the function $f(x)=2 \tan x+(2 a+1) \log _e|\sec x|+(a-2) x$ is increasing on R, then

$a=1 / 2$

If $f(x)=2 \tan x+(2 a+1) \log _e|\sec x|+(a-2) x$ is increasing on R, then

$f'(x) \geq 0$ for all $x \in R$

$\Rightarrow 2 \sec ^2 x+(2 a+1) \tan x+(a-2) \geq 0$ for all $x \in R$

$\Rightarrow 2 \tan ^2 x+(2 a+1) \tan x+a \geq 0$ for all $x \in R$

$\Rightarrow (2 a+1)^2-8 a \leq 0 \Rightarrow(2 a-1)^2 \leq 0 \Rightarrow a=\frac{1}{2}$