Answer
$-0.3218$
Work Step by Step
The Taylor approximation for a function $\tan^{-1} x$ centred at $a=0$ can be written as:
$p(x)=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-.........$----
Now, we have:
$p(\dfrac{-1}{3})=\dfrac{-1}{3}-\dfrac{(\dfrac{-1}{3})^3}{3}+\dfrac{(\dfrac{-1}{3})^5}{5}-.......=-0.3218$
Therefore, $\tan^{-1}(\dfrac{-1}{3})=-0.3218$
