Answer
$$T_{3}(x) =\frac{\pi}{4}+\frac{1}{2}(x-1)-\frac{1}{4}(x-1)^{2}+\frac{1}{12}(x-1)^{3} $$
Work Step by Step
Given $$f(x) =\tan^{-1}x,\ \ a=1$$
Since
\begin{align*}
f(x) &=\tan^{-1}x\ \ \ \ \ \ \ \ \ \ f(1) =\frac{\pi}{4}\\
f'(x) &=\frac{ 1}{1+x^2}\ \ \ \ \ \ \ f'(1 )=\frac{1}{2}\\
f''(x) &=\frac{-2x}{(1+x^2)^2}\ \ \ \ \ \ \ f''(1 )=\frac{-1}{2}\\
f'''(x) &=\frac{6x^2-2}{(1+x^2)^2}\ \ \ \ \ \ \ f'''(1 )=\frac{1}{2}
\end{align*}
Then
\begin{aligned}
T_{3}(x) &=\sum_{n=0}^{3} \frac{f^{(n)}(1)}{n !}(x-1)^{n}\\
&=\frac{\pi}{4}+\frac{1 / 2}{1}(x-1)^{1}+\frac{-1 / 2}{2}(x-1)^{2}+\frac{1 / 2}{6}(x-1)^{3} \\
&=\frac{\pi}{4}+\frac{1}{2}(x-1)-\frac{1}{4}(x-1)^{2}+\frac{1}{12}(x-1)^{3} \end{aligned}
