If f(x)=x+tanx and f is inverse of g, then g1(x) is equal to

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

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

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

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

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

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

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

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

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