The expression to simplify is:

$\cot^{-1} \left( \frac{-1}{\sqrt{x^2 - 1}} \right), \quad \text{for} \quad x < -1.$

Steps to Simplify:

1. Recognize that $\cot^{-1}(y)$ is the inverse cotangent, or arccotangent, of $y$. So, we need to simplify the expression inside the inverse cotangent: $\frac{-1}{\sqrt{x^2 - 1}}.$

2. Let’s substitute this into a trigonometric identity that links it to a simpler form.

   Notice that $\frac{-1}{\sqrt{x^2 - 1}}$ is similar to the form of the expression for the hyperbolic cotangent, but since we're working with inverse cotangent, we use: $\cot^{-1} \left( \frac{-1}{\sqrt{x^2 - 1}} \right) = \tan^{-1}(\sqrt{x^2 - 1}),$ because the relationship between $\cot^{-1} y$ and $\tan^{-1} y$ is: $\cot^{-1}(y) = \tan^{-1} \left( \frac{1}{y} \right).$

3. Finally, this simplifies the expression to: $\tan^{-1} (\sqrt{x^2 - 1}).$

Thus, the simplified expression is:

$\boxed{\tan^{-1}(\sqrt{x^2 - 1})}.$

Would you like further details on this simplification?

Related Questions:

1. What are the conditions under which $\cot^{-1}(x)$ and $\tan^{-1}(x)$ are related?
2. How do the domains of the inverse cotangent and inverse tangent functions affect simplification?
3. What are the other common trigonometric identities used for inverse functions?
4. How does the behavior of $\sqrt{x^2 - 1}$ change for different values of $x$?
5. How do hyperbolic trigonometric functions relate to standard trigonometric functions?

Tip: Always check the domain of a function when simplifying inverse trigonometric expressions, as it affects the range and possible transformations.