Answer
$ x^2-\dfrac{2}{3}x^4+\dfrac{23}{45}x^6-\dfrac{44}{105}x^8+.....$
Work Step by Step
The Taylor series for $\tan^{-1} x $ at $ x=0$ can be defined as: $\tan^{-1} x= x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....$
Consider the given series:
$(\tan^{-1} x)^2= (x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....) \times (x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....) $
or, $=x^2+(-\dfrac{1}{3}-\dfrac{1}{3}) x^4+(\dfrac{1}{5}+\dfrac{1}{(3)(3)}+\dfrac{1}{5}) x^6+......$
or, $=x^2-\dfrac{2}{3}x^4+\dfrac{23}{45}x^6-\dfrac{44}{105}x^8+.....$
