We have been taught that $\csc^{-1}(\csc 0)$ is undefined, as $\csc 0$ is not defined. But when I graphed it on desmos, $\csc^{-1}(\csc x)$ was defined on $x = 0$ and was equal to $0$.
Same is the case for $\sec^{-1}(\sec x)$ at $x = \frac{\pi}{2}$. Can someone explain this discrepancy? Thank you.
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These kinds of identities, where a function that would normally not be defined at a certain point is instead defined as a particular value, show removable singularities. $\lim_{x\rightarrow0}\csc^{-1}\csc x$ is 0 and $\lim_{x\rightarrow\pi/2}\sec^{-1}\sec x$ is $\pi/2$; by defining the values of these functions at these removable singularities as these limits, the functions become regular around those points.
For $x\in [-\pi/2, \pi/2]$, we have $\csc^{-1}(\csc(x))=x$ except when $x=0$, where it is undefined. The reason the graph makes it look like it is defined at $0$ is because you can't see one point being excluded from a curve, unless it is intentionally labeled.
As $x$ approaches $0$, $\csc(x)$ approaches $\pm \infty$, and so $\csc^{-1}(\csc(x))$ approaches $0$.