Solve `tan^(2) theta+cot^(2) theta=2`.

Explanation

To solve the equation \( \tan^2 \theta + \cot^2 \theta = 2 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \tan^2 \theta + \cot^2 \theta = 2 \]
We can rewrite \( \cot^2 \theta \) in terms of \( \tan^2 \theta \):
\[ \cot^2 \theta = \frac{1}{\tan^2 \theta} \]
Substituting this into the equation gives:
\[ \tan^2 \theta + \frac{1}{\tan^2 \theta} = 2 \]

### Step 2: Let \( x = \tan^2 \theta \)
Let \( x = \tan^2 \theta \). Then, the equation becomes:
\[ x + \frac{1}{x} = 2 \]

### Step 3: Multiply through by \( x \)
To eliminate the fraction, multiply both sides by \( x \) (assuming \( x \neq 0 \)):
\[ x^2 + 1 = 2x \]

### Step 4: Rearrange the equation
Rearranging gives us a standard quadratic equation:
\[ x^2 - 2x + 1 = 0 \]

### Step 5: Factor the quadratic
The quadratic can be factored as:
\[ (x - 1)^2 = 0 \]

### Step 6: Solve for \( x \)
Setting the factored equation to zero gives:
\[ x - 1 = 0 \implies x = 1 \]
Since \( x = \tan^2 \theta \), we have:
\[ \tan^2 \theta = 1 \]

### Step 7: Solve for \( \theta \)
Taking the square root gives:
\[ \tan \theta = 1 \quad \text{or} \quad \tan \theta = -1 \]
The general solutions for \( \theta \) are:
\[ \theta = \frac{\pi}{4} + n\pi \quad \text{where } n \in \mathbb{Z} \]
This accounts for both positive and negative values of the tangent function.

### Final Answer
Thus, the solution to the equation \( \tan^2 \theta + \cot^2 \theta = 2 \) is:
\[ \theta = \frac{\pi}{4} + n\pi \quad \text{where } n \in \mathbb{Z} \]