If $\int \frac{2\sec x\tan x}{{(\sec x-\tan x)}^{10}}dx=(\sec x+\tan x){}^{11}[m{(\sec x-tanx)}^2+n]+c$  then $(\frac{1}{m}+\frac{1}{n})$ is equal to (where ‘c’ is constant of integration)

$\int \frac{2\sec x\tan x}{{(\sec x-\tan x)}^{10}}dx$

Let $\sec x-\tan x=t$

$\frac{{}_{+}\mathrm{s}ecx+\tan x=\frac{1}{t}}{2\sec x=t+\frac{1}{t}}$

$\left(2\sec x\tan x\right)dx=(1-\frac{1}{t^2})dt$

$I=\int \frac{(1-\frac{1}{t^2})dt}{t^{10}}$

$I=-\frac{1}{9t^9}+\frac{1}{11t^{11}}+c$

$I={(\sec x+\tan x)}^{11}[-\frac{1}{9}{(\sec x-\tan x)}^2+\frac{1}{11}]+c$