If f(x) = x + tan x and g(x) is the inverse of f(x) then g'(x) is equal to

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

We have

f(x) = x + tan x

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

$\Rightarrow x=g(x)+\tan (g(x)) \quad \left[∵ g(x)=f^{-1}(x)\right]$........(i)

$\Rightarrow 1=g'(x)+\sec ^2(g(x)) g'(x)$

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

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

$\Rightarrow g'(x)=\frac{1}{2+[x-g(x)]^2}$                  [Using (i)]