#### Answer

$L(x)= 1-x$

#### Work Step by Step

The linear approximation of $f(x)$ at $x=0$ is given as:
$L(x)=f(0)+x f'(x)$
We are given that
$f(x)=\dfrac{1}{1+\tan x}$ and $f(0)=\dfrac{1}{1+\tan(0)}=0$
Also,
$f'(x)=\dfrac{-\sec^2 x}{(1+\tan x)^2} \implies f'(0)=\dfrac{-\sec^2 0}{(1+\tan 0)^2}=-1$
Hence, $L(x)=1+x(-1) \implies 1-x$
Hence, the result has been proved.