If $f^{\prime \prime }\left(x\right)>0\mathrm{\forall }x\in R,f^{\mathrm{\prime }}\left(3\right)=0,$ and $g\left(x\right)=f({\tan }^{2}x-2\tan x+4),0<x<\frac{\pi }{2},$ then $g\left(x\right)$ is increasing in

$g^{\mathrm{\prime }}\left(x\right)=\left(f\left(\right(\tan x-1{)}^2+3)\right)2(\tan x-1){\sec }^{2}x$

Since $f^{\mathrm{\prime }}\left(x\right)>0,f^{\mathrm{\prime }}\left(x\right)$ is increasing so,

$f^{\mathrm{\prime }}\left(\right(\tan x-1{)}^2+3)>f^{\mathrm{\prime }}\left(3\right)=0\mathrm{\forall }x\in (0,\frac{\pi }{4})\cup (\frac{\pi }{4},\frac{\pi }{2})$

Also, $(\tan x-1)>0x\in (\frac{\pi }{4},\frac{\pi }{2})$,

so, $g\left(x\right)$ is increasing in $(\frac{\pi }{4},\frac{\pi }{2})$