Inverse related inverse trigonometric functions Prove that for $x<0$,
$\sin^{-1}\sqrt{1-x^2}=\cos^{-1}(-x)$
This question is related to Evaluate ${\cos ^{ - 1}}\sqrt {1 - {x^2}} = $ for $x<0$ but want to solve it differently.
I checked it on

If I use $x=\cos\theta$ then i get $\sin^{-1}|\sin\theta|$ but not able to use the sign
 A: You can let $\arcsin(\sqrt{1-x^{2}}) = \theta \implies \sin(\theta) = \frac{\sqrt{1-x^{2}}}{1}$.
Now you can make a right-angled triangle with the hypotenuse, base and height as $1, -x$ and $\sqrt{1-x^{2}}$ respectively. (Pythagorus' theorem)
When trying to find the base in the right-angled triangle, you will get $y=-x$ and $y=x$ but since $x<0$, we choose $y=-x$, since length cannot be negative.
After this you can find $\cos(\theta)$ easily from which the result follows.
A: Let $\arccos(-x) = \theta$.  Then $\theta$ is the unique angle in the interval $[0, \pi]$ such that $\cos\theta = -x$.  Since $x < 0$, $-x > 0$, so $\arccos(-x) = \theta \in [0, \pi/2)$.
\begin{align*}
\sin^2\theta + \cos^2\theta & = 1\\
\sin^2\theta & = 1 - \cos^2\theta\\
             & = 1 - (-x)^2\\
             & = 1 - x^2\\
|\sin\theta| & = \sqrt{1 - x^2}\\
\sin\theta & = \sqrt{1 - x^2}
\end{align*}
since $\sin\theta \geq 0$ in the interval $[0, \pi/2)$.
Let $\arcsin \sqrt{1 - x^2} = \varphi$.  Then $\varphi$ is the unique angle in the interval $[-\pi/2, \pi/2]$ such that $\sin\varphi = \sqrt{1 - x^2}$.  The only such angle is $\theta$.  Hence, $\varphi = \theta$.  Thus, $\arcsin \sqrt{1 - x^2} = \arccos(-x)$.
