Yes! You can apply the difference of squares formula here. The given expression is:

$\frac{\csc^2(\theta) - \cot^2(\theta)}{\csc(\theta) - \cot(\theta)}$

Since $\csc^2(\theta) - \cot^2(\theta)$ is a difference of squares, you can factor it as:

$(\csc(\theta) - \cot(\theta))(\csc(\theta) + \cot(\theta))$

Now, when you divide by $\csc(\theta) - \cot(\theta)$, that term cancels out, leaving:

$\csc(\theta) + \cot(\theta)$

So, the simplified expression is:

$\csc(\theta) + \cot(\theta)$

Would you like a step-by-step breakdown of the solution?

Here are five related questions for practice:

1. How do you prove that $\csc^2(\theta) - \cot^2(\theta) = 1$?
2. How do you simplify $\frac{x^2 - y^2}{x - y}$ using the difference of squares?
3. What are the fundamental trigonometric identities involving $\csc(\theta)$ and $\cot(\theta)$?
4. How can you express $\csc(\theta)$ and $\cot(\theta)$ in terms of $\sin(\theta)$ and $\cos(\theta)$?
5. What is the result of simplifying $\frac{\sec^2(\theta) - 1}{\tan(\theta)}$?

Tip: The difference of squares formula states that $a^2 - b^2 = (a - b)(a + b)$, which is useful for algebraic simplifications in trigonometry.