https://file.aiquickdraw.com/mathbot/file/17266681105092qd545tx.jpg

Let's solve the equation from the image:

\[
(1 - \sin^2 A)(1 + \cot^2 A) = \cot^2 A
\]

### Step-by-step solution:

1. **Recognize an identity**:
   Notice that \(1 - \sin^2 A\) can be replaced using the Pythagorean identity:
   \[
   1 - \sin^2 A = \cos^2 A
   \]
   So the equation becomes:
   \[
   \cos^2 A \cdot (1 + \cot^2 A) = \cot^2 A
   \]

2. **Substitute the definition of \(\cot^2 A\)**:
   Recall that:
   \[
   \cot^2 A = \frac{\cos^2 A}{\sin^2 A}
   \]
   Now substitute \(\cot^2 A\) into the equation:
   \[
   \cos^2 A \cdot \left(1 + \frac{\cos^2 A}{\sin^2 A}\right) = \frac{\cos^2 A}{\sin^2 A}
   \]

3. **Simplify the expression inside the parentheses**:
   Combine the terms inside the parentheses:
   \[
   1 + \frac{\cos^2 A}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\sin^2 A}
   \]
   Using the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\), this simplifies to:
   \[
   \frac{1}{\sin^2 A}
   \]

4. **Simplify the overall expression**:
   Now the equation becomes:
   \[
   \cos^2 A \cdot \frac{1}{\sin^2 A} = \frac{\cos^2 A}{\sin^2 A}
   \]
   Which simplifies to:
   \[
   \frac{\cos^2 A}{\sin^2 A} = \frac{\cos^2 A}{\sin^2 A}
   \]

Since both sides of the equation are equal, the given equation is **verified**.

---

Do you have any questions about the solution or would you like more details on any step? Here are some related questions to consider:

1. What are the key Pythagorean identities in trigonometry?
2. How do you derive the cotangent function from sine and cosine?
3. Can this type of trigonometric verification be applied to other identities?
4. How can you simplify trigonometric expressions involving multiple identities?
5. How do the sine and cosine functions relate to each other geometrically on the unit circle?

**Tip**: Remember that using identities is a key tool in simplifying trigonometric equations.

Learn how to simplify the trigonometric identity (1 - sin^2A)(1 + cot^2A) = cot^2A using Pythagorean identities and cotangent formulas. Step-by-step solution suitable for Grades 9-12.

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