Let's simplify the given expression:

$\frac{\sin^2(x) - 1}{\sin^2(x) - \sin(x)} = 1 + \csc(x)$

Step 1: Simplify the numerator and denominator

Numerator: $\sin^2(x) - 1$

We recognize that $\sin^2(x) - 1$ is a difference of squares, which can be factored as:

$\sin^2(x) - 1 = (\sin(x) - 1)(\sin(x) + 1)$

Denominator: $\sin^2(x) - \sin(x)$

We can factor out a common factor of $\sin(x)$:

$\sin^2(x) - \sin(x) = \sin(x)(\sin(x) - 1)$

Step 2: Substitute the factored expressions into the original equation

Now the expression becomes:

$\frac{(\sin(x) - 1)(\sin(x) + 1)}{\sin(x)(\sin(x) - 1)}$

Step 3: Cancel out the common factors

We see that $(\sin(x) - 1)$ appears in both the numerator and denominator, so we can cancel it out, provided that $\sin(x) \neq 1$:

$\frac{\sin(x) + 1}{\sin(x)} = 1 + \csc(x)$

Step 4: Simplify the left-hand side

We can now separate the terms in the numerator:

$\frac{\sin(x) + 1}{\sin(x)} = \frac{\sin(x)}{\sin(x)} + \frac{1}{\sin(x)} = 1 + \csc(x)$

Conclusion

We have simplified both sides to:

$1 + \csc(x) = 1 + \csc(x)$

This shows that the original equation is true.

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Would you like more details or have any questions about the solution? Here are some related questions you might find useful:

1. What other trigonometric identities are useful for simplifying expressions like this?
2. How can the fundamental trigonometric identities be applied to equations involving multiple trigonometric functions?
3. What would happen if $\sin(x) = 1$ in this expression?
4. Can this method of simplifying apply to other rational trigonometric expressions?
5. What are some real-world applications where simplifying trigonometric identities is important?

Tip: Always look for common factors in both the numerator and denominator of rational expressions. This is often the key to simplifying complex expressions efficiently.