We are given the expression $(1 + \frac{\tan A}{1 + \cot A})^2$, and we aim to simplify it.

Step-by-step solution:

1. Rewrite $\cot A$ in terms of $\tan A$: $\cot A = \frac{1}{\tan A}$ So, the expression becomes: $\left(1 + \frac{\tan A}{1 + \frac{1}{\tan A}}\right)^2$

2. Simplify the denominator: The denominator $1 + \frac{1}{\tan A}$ can be written as: $\frac{\tan A + 1}{\tan A}$ So, the expression becomes: $\left(1 + \frac{\tan A}{\frac{\tan A + 1}{\tan A}}\right)^2$

3. Simplify the fraction: $\frac{\tan A}{\frac{\tan A + 1}{\tan A}} = \frac{\tan A \cdot \tan A}{\tan A + 1} = \frac{\tan^2 A}{\tan A + 1}$ Thus, the expression now becomes: $\left(1 + \frac{\tan^2 A}{\tan A + 1}\right)^2$

4. Combine terms: We now combine the terms in the parentheses: $1 + \frac{\tan^2 A}{\tan A + 1} = \frac{\tan A + 1 + \tan^2 A}{\tan A + 1}$ So, the expression becomes: $\left(\frac{\tan^2 A + \tan A + 1}{\tan A + 1}\right)^2$

Therefore, the simplified form of the expression is: $\boxed{\left(\frac{\tan^2 A + \tan A + 1}{\tan A + 1}\right)^2}$

Would you like more details on any step or have further questions?

Related Questions:

1. How do you derive trigonometric identities for simplifying expressions?
2. What is the difference between tangent and cotangent?
3. Can you explain how to simplify complex trigonometric fractions?
4. How does the Pythagorean identity $\tan^2 A + 1 = \sec^2 A$ apply here?
5. What is the importance of expressing $\cot A$ in terms of $\tan A$?

Tip:

Always rewrite trigonometric functions in terms of sine and cosine if you find it difficult to simplify them.