Ecstacy of 'e'............

The dots would mean unnecessary higher powers.

$\displaystyle (1+ \tan x)^{\cot x} = e^{\frac{\ln(1+\tan x)}{\tan x}} = e^{1 - \frac{\tan x}{2} + \frac{\tan^2 x}{3} + \dots}$

= $\displaystyle e \bigg(e^{\frac{-\tan x}{2} + \frac{\tan^2 x}{3} + \dots} \bigg)$

= $\displaystyle e \bigg(1 -\frac{\tan x}{2} + \frac{tan^2 x}{3} \dots + \frac{1}{2} \big(-\frac{\tan x}{2} + \frac{tan^2 x}{3} \dots\big)^2 + \dots\bigg)$

= $\displaystyle e\bigg( 1 - \frac{\tan x}{2} + \frac{\tan^2 x}{3} + \frac{\tan^2 x}{8 } + \dots\bigg)$

= $\displaystyle e \bigg(1 - \frac{tan x}{2} + \frac{11}{24} \tan^2 x + \dots\bigg)$

Hence, the numerator of the asked expression becomes $e\bigg(\frac{11}{24} \tan^2 x + \dots \bigg)$

Hence, the answer is $\boxed{\frac{11e}{24}}$