Thank you for the clarification! The problem now reads:

$\text{If } (\tan A + \cot A)^2 = \text{?}$

Let's solve it.

Step 1: Expand the square

$(\tan A + \cot A)^2 = \tan^2 A + 2 \tan A \cdot \cot A + \cot^2 A$

Step 2: Simplify the expression

Recall that $\cot A = \frac{1}{\tan A}$, so we have: $(\tan A + \cot A)^2 = \tan^2 A + 2 \cdot \tan A \cdot \frac{1}{\tan A} + \frac{1}{\tan^2 A}$

Simplify further: $= \tan^2 A + 2 + \frac{1}{\tan^2 A}$

Step 3: Use the identity for $\tan^2 A$

Let $x = \tan^2 A$. Then the expression becomes: $x + 2 + \frac{1}{x}$

Step 4: Common denominator (if needed)

If you need a single expression, you can write this as: $\frac{x^2 + 2x + 1}{x} = \frac{(x + 1)^2}{x}$

But most likely, the answer to the original problem is: $(\tan A + \cot A)^2 = \tan^2 A + 2 + \cot^2 A$

If the question asks for a particular value of $A$, then we would need to substitute that value in.

Would you like more details or have any questions?

Relative Questions:

1. What is the value of $\tan A$ when $A = 45^\circ$?
2. How does the identity $\tan^2 A + 1 = \sec^2 A$ relate to this problem?
3. What happens if $\tan A = 1$ in the original expression?
4. How can this problem be solved using trigonometric identities?
5. What is the geometric interpretation of $\tan A$ and $\cot A$?

Tip:

Understanding the relationship between $\tan$ and $\cot$ functions can greatly simplify trigonometric identities and expressions!