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Q. If $\sin^{-1}\left(\sin\left(\frac{2x^2+4}{1+x^2}\right)\right)<\pi-3$, then $x$ belongs to

Inverse Trigonometric Functions

Solution:

$sin^{-1}\left\{sin\left(\frac{2x^{2}+4}{1+x^{2}}\right)\right\}< \pi -3$
$\Rightarrow sin^{-1} \left(sin\, t\right) < \pi -3$ where $ t $
$ =\frac{ 2\,x^{2} +4}{1+x^{2}} = 2 + \frac{2}{1+x^{2}}$
$ \Rightarrow \pi -t < \pi-3 $
$ \Rightarrow t > 3$
$ \Rightarrow 2+ \frac{2}{1+x^{2}} > 3 $
$\Rightarrow \frac{2}{1+x^{2}} >1 $
$\Rightarrow 2 > 1 + x^{2}$
$ \Rightarrow x^{2}-1 < 0 $
$ \Rightarrow \left|x\right|^{2} <1$
$ \Rightarrow \left|x\right| <1 $
$ \Rightarrow -1 < x < 1$
$ \Rightarrow x \in \left(-1, 1 \right)$