If `tan alpha + cot alpha = 2`, then `tan^(2) alpha + cot^(2) alpha` is

Options

A. 2
B. 16
C. 64
D. 128

Answer

To solve the problem where \( \tan \alpha + \cot \alpha = 2 \) and we need to find \( \tan^2 \alpha + \cot^2 \alpha \), we can follow these steps:

### Step 1: Square both sides of the equation
Starting from the given equation:
\[ \tan \alpha + \cot \alpha = 2 \]
We square both sides:
\[ (\tan \alpha + \cot \alpha)^2 = 2^2 \]

### Step 2: Expand the left-hand side
Using the identity \( (A + B)^2 = A^2 + B^2 + 2AB \), we can expand the left-hand side:
\[ \tan^2 \alpha + \cot^2 \alpha + 2 \tan \alpha \cot \alpha = 4 \]

### Step 3: Substitute the value of \( \tan \alpha \cot \alpha \)
We know that:
\[ \tan \alpha \cot \alpha = 1 \]
Thus, substituting this into the equation gives:
\[ \tan^2 \alpha + \cot^2 \alpha + 2 \cdot 1 = 4 \]
This simplifies to:
\[ \tan^2 \alpha + \cot^2 \alpha + 2 = 4 \]

### Step 4: Solve for \( \tan^2 \alpha + \cot^2 \alpha \)
Now, we can isolate \( \tan^2 \alpha + \cot^2 \alpha \):
\[ \tan^2 \alpha + \cot^2 \alpha = 4 - 2 \]
\[ \tan^2 \alpha + \cot^2 \alpha = 2 \]

### Final Answer
Thus, the value of \( \tan^2 \alpha + \cot^2 \alpha \) is:
\[ \boxed{2} \]