We are tasked with simplifying the expression:

\[
\coth(x) + \coth(x) \csc^2(x)
\]

### Step 1: Factor \(\coth(x)\)
We can factor out \(\coth(x)\) from both terms:

\[
\coth(x) + \coth(x) \csc^2(x) = \coth(x) \left(1 + \csc^2(x)\right)
\]

### Step 2: Use a known identity
Recall that \(\coth(x)\) is defined as:

\[
\coth(x) = \frac{\cosh(x)}{\sinh(x)}
\]

and \(\csc(x)\) is related to sine:

\[
\csc(x) = \frac{1}{\sin(x)}
\]

However, since there is no direct identity involving both coth and \(\csc^2(x)\) that simplifies further in this form, the expression is already factored and simplified.

Thus, the simplified form of the given expression is:

\[
\boxed{\coth(x) \left(1 + \csc^2(x)\right)}
\]

Would you like more details, or do you have any other questions?

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Here are some related questions you might want to explore:
1. What is the derivative of \(\coth(x)\)?
2. How is \(\csc(x)\) related to the other trigonometric functions?
3. What is the integral of \(\coth(x)\)?
4. How can hyperbolic functions be used in real-world applications?
5. Can you simplify \(\csch(x) + \csch(x)\csc(x)\)?

**Tip:** Hyperbolic functions like \(\coth(x)\) are analogous to trigonometric functions and can be useful in many areas of mathematics, especially in calculus and differential equations.

Learn how to simplify the expression coth(x) + coth(x)csc^2(x) using algebraic factoring and trigonometric identities. This problem involves hyperbolic functions like coth and trigonometric functions like csc(x), making it suitable for advanced high school students or early college learners.

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