Answer
$$L(x)= x,\ \ \ 0.053\%$$
Work Step by Step
Given $$\tan(0.04)$$
Consider $f(x)=\tan x, a=0,$ and $\Delta x=0.04$, since
\begin{align*}
f^{\prime}(x)&=\sec^2 x\\
f^{\prime}(0)&= 1
\end{align*}
Then the linearization to $f (x)$ is given by
\begin{align*}
L(x)&=f^{\prime}(a)(x-a)+f(a)\\
&=x
\end{align*}
Since
\begin{align*}
L(0.04) &=(0.04)
\end{align*}
Hence the error is given by
$$ |\tan(0.04)-0.04|=0.039301$$
and the percentage is
$$\frac{0.039301 }{\tan(0.04)}\times 100 \% \approx 0.053\%$$
