Let $f(x)=\frac{e^{\tan x}-e^x+\ln (\sec x+\tan x)-x}{\tan x-x}$ be continuous at x = 0. Then the value of f(0), is

$\frac{3}{2}$

For f(x) to be continuous at x = 0, we must have

$\lim\limits_{x \rightarrow 0} \frac{e^{\tan x}-e^x+\ln (\sec x+\tan x)-x}{\tan x-x}=f(0)$

$\Rightarrow f(0)=\lim\limits_{x \rightarrow 0} e^x \frac{\left(e^{\tan x-x}-1\right)}{\tan x-x}+\lim\limits_{x \rightarrow 0} \frac{\ln (\sec x+\tan x)-x}{\tan x-x}$

$\Rightarrow f(0)=\lim\limits_{x \rightarrow 0} e^x\left(\frac{e^{\tan x-x}-1}{\tan x-x}\right)+\lim\limits_{x \rightarrow 0} \frac{\log (\sec x+\tan x)-x}{x^3\left(\frac{\tan x-x}{x^3}\right)}$

$\Rightarrow f(0)=1 \times e^0+3 \lim\limits_{x \rightarrow 0} \frac{\ln (\sec x+\tan x)-x}{x^3}$              $\left[∵ \lim\limits_{x \rightarrow 0} \frac{\tan x-x}{x^3}=\frac{1}{3}\right]$

$\Rightarrow f(0)=1+3 \lim\limits_{x \rightarrow 0} \frac{\sec x-1}{3 x^2}$               [By L' Hospital's rule]

$\Rightarrow f(0)=1+3 \lim\limits_{x \rightarrow 0} \frac{1-\cos x}{3 \cos x . x^2}$

$\Rightarrow f(0)=1+3 \times \frac{1}{3} \times \frac{1}{2}=\frac{3}{2}$