Answer
$$ x^2-\dfrac{2}{3}x^4+\dfrac{23}{45}x^6-\dfrac{44}{105}x^8+.....$$
Work Step by Step
Recall the Taylor series for: $\tan^{-1} x= x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....$
$(\tan^{-1} x)^2= (x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....) ^2= (x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....) \times (x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....) $
or, $(\tan^{-1} x)^2=x^2+(-\dfrac{1}{3}-\dfrac{1}{3}) x^4+(\dfrac{1}{5}+\dfrac{1}{(3)(3)}+\dfrac{1}{5}) x^6+......$
or, $(\tan^{-1} x)^2=x^2-\dfrac{2}{3}x^4+\dfrac{23}{45}x^6-\dfrac{44}{105}x^8+.....$
