Answer
See below.
Work Step by Step
$\sin^{-1}x$ is the number $y $in $[-\pi/2,\pi/2]$ for which $\sin y=x$
$\cos^{-1}x$ is the number $y $in $[0,\pi]$ for which $\cos y=x$
$\sin^{-1}-1=-\pi/2,\quad \cos^{-1}-1=\pi$
... so the equation stands for $x=-1$
$\sin^{-1}0=0,\quad \cos^{-1}0=\pi/2$
... so the equation stands for $x=0$
$\sin^{-1}1=\pi/2,\quad \cos^{-1}1=0$
... so the equation stands for $x=1$
For $a\in(-1,0)$, its opposite, $x=-a$ is in $(0,1)$
equation (1) $\Rightarrow \sin^{-1}(a)=\sin^{-1}(-x)=-\sin^{-1}x$
equation (3) $\Rightarrow \cos^{-1}(a)=\pi-\cos^{-1}x$
Adding the two,
$\sin^{-1}(a)+\cos^{-1}(a)=\pi-(\sin^{-1}x+\cos^{-1}x)$
... now, by equation (4), the sum in the parentheses equals $\pi/2,$ so
$\sin^{-1}(a)+\cos^{-1}(a)=\pi-\pi/2=\pi/2$
... so, the equation is valid for values in $[-1,0]$ as well.
