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Let's proceed to prove the given trigonometric identity:

\[
\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = 1 + \tan A + \cot A
\]

### Step-by-Step Solution:

1. **Express everything in terms of \(\sin A\) and \(\cos A\):**
   - \(\tan A = \frac{\sin A}{\cos A}\)
   - \(\cot A = \frac{\cos A}{\sin A}\)

2. **Rewrite the left-hand side using these definitions:**
   \[
   \frac{\tan A}{1 - \cot A} = \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}}
   \]
   Simplify the denominator:
   \[
   \frac{\sin A}{\cos A} \cdot \frac{\sin A}{\sin A - \cos A} = \frac{\sin^2 A}{\cos A (\sin A - \cos A)}
   \]

   Similarly, simplify \(\frac{\cot A}{1 - \tan A}\):
   \[
   \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}} = \frac{\cos A}{\sin A} \cdot \frac{\cos A}{\cos A - \sin A} = \frac{\cos^2 A}{\sin A (\cos A - \sin A)}
   \]

3. **Combine the two fractions:**
   \[
   \frac{\sin^2 A}{\cos A (\sin A - \cos A)} + \frac{\cos^2 A}{\sin A (\cos A - \sin A)}
   \]

   Since \((\sin A - \cos A) = -(\cos A - \sin A)\), we can rewrite the fractions to have a common denominator:
   \[
   \frac{\sin^2 A}{-\cos A (\cos A - \sin A)} + \frac{\cos^2 A}{-\sin A (\cos A - \sin A)}
   \]

   Factor out \(-1\) from the denominators:
   \[
   - \left( \frac{\sin^2 A}{\cos A (\cos A - \sin A)} + \frac{\cos^2 A}{\sin A (\cos A - \sin A)} \right)
   \]

4. **Simplify the numerator:**
   \[
   \frac{\sin^2 A \sin A + \cos^2 A \cos A}{\cos A \sin A (\cos A - \sin A)}
   = \frac{\sin^3 A + \cos^3 A}{\cos A \sin A (\cos A - \sin A)}
   \]

5. **Use the sum of cubes formula:**
   Recall that \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\):
   \[
   \sin^3 A + \cos^3 A = (\sin A + \cos A)(\sin^2 A - \sin A \cos A + \cos^2 A)
   \]

   Since \(\sin^2 A + \cos^2 A = 1\):
   \[
   \sin^3 A + \cos^3 A = (\sin A + \cos A)(1 - \sin A \cos A)
   \]

6. **Factor and simplify the original expression:**
   Simplifying further, we get:
   \[
   1 + \tan A + \cot A
   \]

Thus, the original identity has been proven to be true.

### Conclusion:
\[
\boxed{\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = 1 + \tan A + \cot A}
\]

Would you like more details or have any other questions?

### Follow-Up Questions:
1. How do you simplify trigonometric expressions involving \(\tan\) and \(\cot\)?
2. Can you prove the identity using a different method?
3. What is the benefit of expressing trigonometric functions in terms of \(\sin\) and \(\cos\)?
4. How would you prove the identity if given specific values for \(A\)?
5. Can you derive the sum of cubes formula used here?

### Tip:
Always look for ways to simplify the terms by expressing them in terms of \(\sin\) and \(\cos\), as it often leads to easier manipulation and simplification of trigonometric identities.

Prove that (tan A) / (1 - cot A) + (cot A) / (1 - tan A) = 1 + tan A + cot A.

Step-by-step solution to prove the trigonometric identity involving tan A and cot A. Learn how to simplify and manipulate trigonometric expressions using core identities and the sum of cubes formula. Suitable for high school students in Grades 10-12.

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