We aim to simplify the expression $\cos(\beta)\tan(\beta)\csc(\beta)\cot(\beta)$.

Step 1: Recall trigonometric identities

1. $\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}$,
2. $\csc(\beta) = \frac{1}{\sin(\beta)}$,
3. $\cot(\beta) = \frac{\cos(\beta)}{\sin(\beta)}$.

Substituting these into the expression:

$\cos(\beta) \tan(\beta) \csc(\beta) \cot(\beta) = \cos(\beta) \cdot \frac{\sin(\beta)}{\cos(\beta)} \cdot \frac{1}{\sin(\beta)} \cdot \frac{\cos(\beta)}{\sin(\beta)}.$

Step 2: Simplify the expression

Cancel terms wherever possible:

1. $\frac{\sin(\beta)}{\sin(\beta)} = 1$,
2. $\frac{\cos(\beta)}{\cos(\beta)} = 1$.

Thus, the expression simplifies to:

$\frac{\cos(\beta)}{\sin(\beta)} = \cot(\beta).$

Step 3: Determine the domain

For the expression to be defined:

1. $\cos(\beta) \neq 0$ to avoid division by zero in $\tan(\beta)$ and $\cot(\beta)$,
2. $\sin(\beta) \neq 0$ to avoid division by zero in $\csc(\beta)$ and $\cot(\beta)$.

The domain is therefore:

$\beta \neq n\pi \quad \text{and} \quad \beta \neq \frac{\pi}{2} + n\pi, \quad \text{where } n \text{ is an integer}.$

Final Answer:

$\cos(\beta)\tan(\beta)\csc(\beta)\cot(\beta) = \cot(\beta), \quad \text{defined for } \beta \notin \{n\pi, \frac{\pi}{2} + n\pi\}, \text{where } n \in \mathbb{Z}.$

Would you like a breakdown of a specific step or further clarification?

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Related Questions:

1. How can we simplify $\sin(\beta)\csc(\beta) + \cos^2(\beta)$?
2. What is the domain of $\tan(\beta)\cot(\beta)$?
3. Can $\cot(\beta)$ be rewritten in terms of only $\sin(\beta)$ and $\cos(\beta)$?
4. How does $\tan(\beta)$ behave near $\beta = \frac{\pi}{2}$?
5. What are the key trigonometric identities used for simplifying expressions?

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Tip:

Always check the domain of the original expression before simplifying, as simplification can obscure undefined points.