To determine which function passes through the point [tex]\(\left(0, \frac{\pi}{2}\right)\)[/tex], we need to evaluate each of the given inverse trigonometric functions at [tex]\(x = 0\)[/tex] and see which one results in [tex]\(\frac{\pi}{2}\)[/tex].
### Step-by-Step Solution:
1. Evaluate [tex]\(y = \cos^{-1}(x)\)[/tex] at [tex]\(x = 0\)[/tex]:
[tex]\[
y = \cos^{-1}(0)
\][/tex]
The value of [tex]\(\cos^{-1}(0)\)[/tex] is [tex]\(\frac{\pi}{2}\)[/tex]. Thus, [tex]\((0, \frac{\pi}{2})\)[/tex] lies on the graph of [tex]\(y = \cos^{-1}(x)\)[/tex].
2. Evaluate [tex]\(y = \csc^{-1}(x)\)[/tex] at [tex]\(x = 0\)[/tex]:
[tex]\[
y = \csc^{-1}(0)
\][/tex]
The value [tex]\(\csc^{-1}(0)\)[/tex] is undefined because [tex]\(\csc(x) = \frac{1}{\sin(x)}\)[/tex] is undefined for [tex]\(x = 0\)[/tex]. Hence, [tex]\((0, \frac{\pi}{2})\)[/tex] does not lie on the graph of [tex]\(y = \csc^{-1}(x)\)[/tex].
3. Evaluate [tex]\(y = \sec^{-1}(x)\)[/tex] at [tex]\(x = 0\)[/tex]:
[tex]\[
y = \sec^{-1}(0)
\][/tex]
The value [tex]\(\sec^{-1}(0)\)[/tex] is also undefined because [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex] is undefined for [tex]\(x = 0\)[/tex]. So, [tex]\((0, \frac{\pi}{2})\)[/tex] does not lie on the graph of [tex]\(y = \sec^{-1}(x)\)[/tex].
4. Evaluate [tex]\(y = \sin^{-1}(x)\)[/tex] at [tex]\(x = 0\)[/tex]:
[tex]\[
y = \sin^{-1}(0)
\][/tex]
The value of [tex]\(\sin^{-1}(0)\)[/tex] is 0. Thus, [tex]\((0, 0)\)[/tex] lies on the graph of [tex]\(y = \sin^{-1}(x)\)[/tex]. Therefore, [tex]\((0, \frac{\pi}{2})\)[/tex] does not lie on the graph of [tex]\(y = \sin^{-1}(x)\)[/tex].
### Conclusion:
After evaluating each function at [tex]\(x = 0\)[/tex], we see that only [tex]\(y = \cos^{-1}(x)\)[/tex] satisfies [tex]\((0, \frac{\pi}{2})\)[/tex]. Therefore, the graph of the function [tex]\(y = \cos^{-1}(x)\)[/tex] passes through the point [tex]\(\left(0, \frac{\pi}{2}\right)\)[/tex].
