https://file.aiquickdraw.com/mathbot/file/1740972397543yvbepwfk.jpg

The image you uploaded shows a proof for the identity:

\[
(\csc^2 x - 1) \sec^2 x = \csc^2 x
\]

The goal is to fill in the appropriate rules to justify each step in the proof. Here's how the steps proceed logically:

1. The original expression is \((\csc^2 x - 1) \sec^2 x\).
2. The next step uses the **Pythagorean Identity**: \(\csc^2 x - 1 = \cot^2 x\). This converts the expression into \(\cot^2 x \sec^2 x\).
3. Then, using the **Reciprocal Identity** for \(\cot x = \frac{\cos x}{\sin x}\) and \(\sec x = \frac{1}{\cos x}\), the expression becomes:
   \[
   \frac{\cos^2 x}{\sin^2 x} \cdot \frac{1}{\cos^2 x}
   \]
4. Simplifying the expression, the \(\cos^2 x\) terms cancel, leaving:
   \[
   \frac{1}{\sin^2 x}
   \]
5. Finally, using the **Reciprocal Identity** for \(\csc^2 x = \frac{1}{\sin^2 x}\), the expression is equal to \(\csc^2 x\).

To summarize:

1. **Pythagorean Identity**: \(\csc^2 x - 1 = \cot^2 x\)
2. **Reciprocal Identity** for \(\cot x\) and \(\sec x\)
3. **Simplification**
4. **Reciprocal Identity** for \(\csc^2 x\)

Let me know if you would like further details or clarifications! Would you like to explore similar problems or discuss any specific rule? 

Here are some related questions you may find helpful:

1. What are the key Pythagorean identities for trigonometric functions?
2. How do reciprocal identities work in trigonometry?
3. Can you prove other trigonometric identities using the Pythagorean identity?
4. How do you simplify expressions involving \(\cot x\) and \(\sec x\)?
5. How do algebraic manipulations help in simplifying trigonometric expressions?

**Tip**: Always remember that simplifying trigonometric expressions often involves using identities like Pythagorean and Reciprocal, and simplifying algebraically to cancel terms when possible.

Learn how to prove the trigonometric identity (csc^2 x - 1) sec^2 x = csc^2 x. This proof involves using Pythagorean identities, reciprocal identities, and algebraic simplification. Suitable for students in Grades 10-12.

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