Let for $x\in (0,\frac{\pi }{2}),f\left(x\right)={\cot }^{-1}(2-\tan x)$ and $g\left(\theta \right)={\int }_{\min \{\theta ,\pi /4\}}^{\max \{\theta ,\pi /4\}}f\left(x\right)dx$ ; and the value of $\left(\frac{{\pi }^2}{g\left(\frac{\pi }{12}\right)-g\left(\frac{\pi }{3}\right)}\right)$ is k. The sum of digits of ‘k’ is_______

$g\left(\frac{\pi }{12}\right)=\underset{\pi /12}{\overset{\pi /4}{\int }}{\cot }^{-1}(2-\tan x)dx$

$=x{\cot }^{-1}{(2-\tan x)|}_{\pi /12}^{\pi /4}-\underset{\pi /12}{\overset{\pi /4}{\int }}\frac{x{\sec }^{2}x}{1+{(2-\tan x)}^2}dx$

$=(\frac{\pi }{4}\times \frac{\pi }{4}-\frac{\pi }{12}\times \frac{\pi }{6})-I$

In I ; put $2-\tan x=\tan \theta$  then, $g\left(\frac{\pi }{12}\right)=\frac{7{\pi }^2}{144}-\underset{\pi /4}{\overset{\pi /3}{\int }}(\frac{\pi }{2}-{\cot }^{-1}(2-\tan \theta ))d\theta =\frac{{\pi }^2}{144}+g\left(\frac{\pi }{3}\right)$