We are asked to simplify the expression:

$\frac{\cot(A) - 1}{1 + \tan(-A)}$

Step 1: Apply trigonometric identities

• Recall that $\tan(-A) = -\tan(A)$ and $\cot(A) = \frac{1}{\tan(A)}$.
• Substituting these identities, the expression becomes:

$\frac{\cot(A) - 1}{1 + \tan(-A)} = \frac{\frac{1}{\tan(A)} - 1}{1 - \tan(A)}$

Step 2: Simplify the numerator

• To combine the terms in the numerator, we write $1$ as $\frac{\tan(A)}{\tan(A)}$:

$\frac{1}{\tan(A)} - 1 = \frac{1}{\tan(A)} - \frac{\tan(A)}{\tan(A)} = \frac{1 - \tan^2(A)}{\tan(A)}$

Step 3: Substitute and simplify the entire expression

Now, substitute the simplified numerator back into the expression:

$\frac{\frac{1 - \tan^2(A)}{\tan(A)}}{1 - \tan(A)}$

Step 4: Final simplification

We can rewrite this as:

$\frac{1 - \tan^2(A)}{\tan(A)(1 - \tan(A))}$

This is the simplified form of the given expression.

Final Answer:

$\frac{1 - \tan^2(A)}{\tan(A)(1 - \tan(A))}$

---

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How can we simplify $\tan(A + B)$ using trigonometric identities?
2. What are the basic trigonometric identities that help in simplifying expressions?
3. How do you prove that $\cot(A) = \frac{1}{\tan(A)}$?
4. What is the Pythagorean identity and how does it relate to simplification?
5. Can you show how to simplify an expression like $\sin(A) + \cos(A)$?

Tip: Whenever dealing with trigonometric expressions, it's often helpful to express everything in terms of sine and cosine for easier simplification.