If $f$''(x) is a positive function for all $x\in R,f$'(3) = 0 and $\mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left({\tan }^2\right(\mathrm{x})-2\tan (\mathrm{x})+4)$ for $0<\mathrm{x}<\frac{\mathrm{\pi }}{2},$ then the interval in which g(x) is increasing is $\mathrm{\_\_\_\_\_}$

$$\begin{gathered} \mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left({\tan }^2\mathrm{x}-2 \tan \mathrm{x}+4\right) \\ {\mathrm{g}}^1\left(\mathrm{x}\right)={\mathrm{f}}^1\left({\tan }^2\mathrm{x}-2\tan \mathrm{x}+4\right).\left(2 \tan \mathrm{x} {\sec }^2\mathrm{x}-2 {\sec }^2\mathrm{x}\right) \\ ={\mathrm{f}}^1({\left(\tan \mathrm{x}-1\right)}^2+3).2{\sec }^2\mathrm{x}(\tan \mathrm{x}-1) \end{gathered}$$$$\begin{gathered} \mathrm{g}\left(\mathrm{x}\right) \mathrm{is} \mathrm{increasing} \\ \Rightarrow {\mathrm{g}}^1\left(\mathrm{x}\right)>0 \\ {\left(\tan \mathrm{x}-1\right)}^2\ge 0 \\ {\left(\tan \mathrm{x}-1\right)}^2+3\ge 3 \\ {\mathrm{f}}^1({\left(\tan \mathrm{x}-1\right)}^2+3)\ge {\mathrm{f}}^1\left(3\right)\ge 0 \\ \therefore \tan \mathrm{x}-1>0 \\ \tan \mathrm{x}>1 \\ \mathrm{x}>\mathrm{\pi }/4 \end{gathered}$$