$sin^{-1}\begin{Bmatrix}sin \left(\frac{2x^2+4}{x^2+1}\right)\end{Bmatrix} < \pi - 3,$ if 

x ∈ (-1, 1)

We have,

$\frac{2x^2+4}{x^2+1}=\frac{2(x^2+1)+2}{x^2+1}= 2+\frac{2}{x^2+1}$

$∴ 2 < \frac{2x^2+4}{x^2+1}$ ≤ 4 for all x ∈ R

$⇒ \pi - 4 < \pi -\frac{2x^2+4}{x^2+1} < \pi - 2 $ for all x ∈ R

$ -\frac{\pi}{2} < \pi - \frac{2x^2+4}{x^2+1}<\frac{\pi}{2}$ for all x ∈ R

$∴ sin^{-1}\begin{Bmatrix}sin\left(\frac{2x^2+4}{x^2+1}\right)\end{Bmatrix}= sin^{-1} \begin{Bmatrix}sin\left(\pi - \frac{2x^2+4}{x^2+1}\right)\end{Bmatrix}$

$= \pi - \frac{2x^2+4}{x^2-1}$

Hence, $ sin^{-1}\begin{Bmatrix}sin\left(\frac{2x^2+4}{x^2+1}\right)\end{Bmatrix} , \pi - 3 $

$ ⇒ \pi - \frac{2x^2+4}{x^2+1} < \pi - 3 $

$ ⇒ \frac{2x^2+4}{x^2+1} > 3 $

$ ⇒ 2+\frac{2}{x^2+1}  ⇒ 2 > 3 > x^2 + 1 ⇒ x^2 -1 < 0 ⇒ x ∈ (-1, 1)$