Answer
See explanations.
Work Step by Step
Step 1. Letting $f(x)=\frac{1}{1+tan(x)}$, we have $f'(x)=\frac{-sec^2x}{(1+tan(x))^2}$
Step 2. Calculate function values at $x=0$: $f(0)=\frac{1}{1+tan(0)}=1$ and $f'(0)=\frac{-sec^20}{(1+tan(0))^2}=-1$
Step 3. The linearization of $f(x) $at $x=0$ can be written as $L(x)=f(0)+f'(0)(x-0)=1-x$.
