For What Values of $a$ is $\sin^{-1}(\frac{1}{x^2+a})$ Defined for any Real $x$?

Question

I know that $\sin^{-1}(x)$ is defined for $x \in [-1, 1]$, so: $$-1 \le \frac{1}{x^2+a} \le 1$$ $$| x^2 + a |\ge 1$$ How do I continue from there?

Answer 1

If $a\leq 0$ then there are points where $x^{2}+a=0$ so we have to take $a >0$. Now $-1 \leq \frac 1 {x^{2}+a} \leq 1$ iff $|x^{2}+a| \geq 1$. But $x^{2}+a >0$ so we want $x^{2}+a \geq 1$ for every $x$. This is true iff $a \geq 1$.