Let's determine the domain and range of each function.

(A) $y = \tan(x)$

Domain:
The tangent function, $\tan(x)$, is undefined where the cosine function $\cos(x)$ is zero. This occurs at $x = \frac{\pi}{2} + n\pi$, where $n$ is any integer.

Thus, the domain of $y = \tan(x)$ is: $x \in \mathbb{R} \setminus \left\{ \frac{\pi}{2} + n\pi \mid n \in \mathbb{Z} \right\}$ Or in interval notation: $x \in \left( -\infty, \frac{\pi}{2} \right) \cup \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \cup \left( \frac{3\pi}{2}, \frac{5\pi}{2} \right) \cup \dots$

Range:
The range of $y = \tan(x)$ is all real numbers since the tangent function can take any real value: $y \in \mathbb{R}$

(B) $y = \frac{1}{x(x - 1)}$

Domain:
The function is undefined where the denominator is zero, which occurs at $x = 0$ and $x = 1$.

Thus, the domain of $y = \frac{1}{x(x - 1)}$ is: $x \in \mathbb{R} \setminus \{0, 1\}$ Or in interval notation: $x \in (-\infty, 0) \cup (0, 1) \cup (1, \infty)$

Range:
To determine the range, consider the function's behavior as $x$ approaches the values that make the denominator zero and the limits as $x$ approaches infinity. The function can take on all real values except $y = 0$ because there's no $x$ that makes $\frac{1}{x(x-1)} = 0$.

Thus, the range of $y = \frac{1}{x(x - 1)}$ is: $y \in \mathbb{R} \setminus \{0\}$

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Would you like further details on any of these points, or do you have any questions?

Related Questions:

1. What are the domain and range of $y = \cot(x)$?
2. How does the domain of $y = \sec(x)$ compare to $y = \tan(x)$?
3. What is the domain and range of $y = \frac{1}{x^2 + 1}$?
4. How do asymptotes affect the range of rational functions?
5. Can you explain the domain and range of inverse trigonometric functions?

Tip:

When determining the domain of a function, always consider the points where the function might be undefined, such as where a denominator equals zero or where a function inside a square root is negative.