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The given equation is:

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

We can start simplifying the left-hand side of the equation:

1. Recall that \(1 - \sin^2 A = \cos^2 A\) from the Pythagorean identity.
   
   So, the equation becomes:

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

2. Now, use the identity for \(\cot A = \frac{\cos A}{\sin A}\), which gives us:

   \[
   \cot^2 A = \frac{\cos^2 A}{\sin^2 A}
   \]

   So, the equation now is:

   \[
   \cos^2 A \left( 1 + \frac{\cos^2 A}{\sin^2 A} \right) = \frac{\cos^2 A}{\sin^2 A}
   \]

3. Let's simplify the expression inside the parentheses on the left-hand side:

   \[
   1 + \frac{\cos^2 A}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\sin^2 A}
   \]

   Since \(\sin^2 A + \cos^2 A = 1\) (Pythagorean identity), this becomes:

   \[
   \frac{1}{\sin^2 A}
   \]

4. Now, substitute this back into the equation:

   \[
   \cos^2 A \times \frac{1}{\sin^2 A} = \frac{\cos^2 A}{\sin^2 A}
   \]

5. Both sides are now equal, so the equation is verified:

   \[
   \frac{\cos^2 A}{\sin^2 A} = \frac{\cos^2 A}{\sin^2 A}
   \]

Thus, the equation holds true.

Do you have any questions or would you like a deeper explanation?

Here are some related questions you might explore:

1. What is the significance of the Pythagorean identities in trigonometry?
2. How can we derive the cotangent identity from the basic trigonometric functions?
3. Can this identity be used to solve trigonometric equations more efficiently?
4. How do other trigonometric identities (like secant or cosecant) interact with cotangent?
5. What are some real-world applications of trigonometric identities?

**Tip:** Trigonometric identities are very helpful in simplifying complex expressions. Familiarize yourself with the Pythagorean identities, as they are frequently used in various problems.

Learn how to prove the trigonometric identity (1 - sin^2 A)(1 + cot^2 A) = cot^2 A with step-by-step explanations. This problem focuses on using Pythagorean identities and cotangent properties, suitable for students in Grades 9-12.

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