If $x$ takes a non-positive permissible value, then ${\sin }^{-1}x$ will be:

Let ${\sin }^{-1}x=y$. Then $x=\sin y$.Given $x$ is non-positive and permissible, $-1\le x\le 0$.The range of ${\sin }^{-1}x$ is $[-\pi /2,\pi /2]$. For $-1\le x\le 0$, we have $-\pi /2\le y\le 0$.We know ${\cos }^{2}y=1-{\sin }^{2}y=1-x^2$, so $\cos y=\sqrt{1-x^2}$ (since $\cos y\ge 0$ for $y\in [-\pi /2,0]$).Let $-y=\theta$. Since $-\pi /2\le y\le 0$, we have $0\le \theta \le \pi /2$.Then $\cos (-\theta )=\cos \theta =\sqrt{1-x^2}$.Since $\theta$ is in the principal range of ${\cos }^{-1}$, $\theta ={\cos }^{-1}\sqrt{1-x^2}$.Substituting $\theta =-y$:$-y={\cos }^{-1}\sqrt{1-x^2}⟹y=-{\cos }^{-1}\sqrt{1-x^2}$.Thus, ${\sin }^{-1}x=-{\cos }^{-1}\sqrt{1-x^2}$.