Answer
$$0.02,\ \ \ 0.020202,\ \ \ 0.0002027$$
Work Step by Step
Consider $f(x)=\tan x $, $a= \pi/4 $, $\Delta x=0.01$, since
\begin{align*}
f'(x) &= \sec^2 x \\
f'(\pi/4)&= 2
\end{align*}
Then the linear approximation is given by
\begin{align*}
\Delta &f \approx f^{\prime}(a) \Delta x\\
&= (2)(0.01)\\
&= 0.02
\end{align*}
and the actual change is given by
\begin{align*}
\Delta f&=f(a+\Delta x)-f(a)\\
&=f(\pi/4)-f(\pi/4+0.01)\\
& = 0.0202027
\end{align*}
Hence the error is
$$|0.0202027-0.02|=0.0002027$$
