Answer
$$0.025,\ \ \ 0.02438,\ \ 0.00062$$
Work Step by Step
Given $$\tan ^{-1}(1.05)-\frac{\pi}{4} $$
Consider $f(x)= \tan ^{-1} x $, $a= 1$, $\Delta x= 0.05$, since
\begin{align*}
f'(x) &= \frac{1}{1+x^2} \\
f'(1)&=0.5
\end{align*}
Then the linear approximation is given by
\begin{align*}
\Delta &f \approx f^{\prime}(a) \Delta x\\
&= (0.5)(0.05)\\
&= 0.025
\end{align*}
and the actual change is given by
\begin{align*}
\Delta f&=f(a+\Delta x)-f(a)\\
&=f(1.05)-f(1)\\
& = 0.02438
\end{align*}
Hence the error is
$$ |0.02438-0.025| \approx 0.00062 $$
