${\tan }^{-1}(\tan \sqrt{1-\theta })=\sqrt{1-\theta }$, when 

Clearly, $\sqrt{1-0}$ is real, if $\theta \le 1$.

${\tan }^{-1}(\tan \sqrt{1-\theta })=\sqrt{1-\theta }$

$\Rightarrow 0\le \sqrt{1-\theta }<\frac{\pi }{2}$and $\theta \le 1$

$\Rightarrow 0\le 1-\theta <\frac{{\pi }^2}{4}$and $\theta \le 1$

$\Rightarrow -1\le -\theta \le \frac{{\pi }^2}{4}-1$and $\theta \le 1$

$\Rightarrow 1-\frac{{\pi }^2}{4}<\theta \le 1$ and $\theta \le 1\Rightarrow \frac{4-{\pi }^2}{4}<\theta \le 1$