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^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.