If $f\left(x\right)=x+\tan x$ and f is inverse of g, then $g^1\left(x\right)$ is equal to 

$g\left(x\right)=f^{-1}\left(x\right)\Rightarrow f\left(g\left(x\right)\right)=x$

$\Rightarrow f^{|}\left(g\left(x\right)\right)g^{|}\left(x\right)=1$

$\Rightarrow g^{|}\left(x\right)=\frac{1}{f^{|}\left(g\left(x\right)\right)}$

 $=\frac{1}{1+{\sec }^2\left(g\left(x\right)\right)}$

 $=\frac{1}{1+{\tan }^{2}g\left(x\right)+1}$

  $=\frac{1}{2+\left[f\left(g\left(x\right)\right)\right]-{\left(g\left(x\right)\right)}^2}$

 $=\frac{1}{2+{(x-g\left(x\right))}^2}$

 $=\frac{1}{2+{(g\left(x\right)-x)}^2}$