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 given equation:
\[ \tan^2 \theta + \cot^2 \theta = 2 \]
We can rearrange this as:
\[ \tan^2 \theta + \cot^2 \theta - 2 = 0 \]

**Hint:** Recognize that \( \cot \theta = \frac{1}{\tan \theta} \).

### Step 2: Use the identity for cotangent
Since \( \cot \theta = \frac{1}{\tan \theta} \), we can substitute this into the equation:
\[ \tan^2 \theta + \left(\frac{1}{\tan \theta}\right)^2 - 2 = 0 \]
This simplifies to:
\[ \tan^2 \theta + \frac{1}{\tan^2 \theta} - 2 = 0 \]

**Hint:** Consider multiplying through by \( \tan^2 \theta \) to eliminate the fraction.

### Step 3: Multiply through by \( \tan^2 \theta \)
Multiplying the entire equation by \( \tan^2 \theta \) (assuming \( \tan \theta \neq 0 \)):
\[ \tan^4 \theta - 2\tan^2 \theta + 1 = 0 \]

**Hint:** This is a quadratic equation in terms of \( x = \tan^2 \theta \).

### Step 4: Let \( x = \tan^2 \theta \)
Let \( x = \tan^2 \theta \). The equation becomes:
\[ x^2 - 2x + 1 = 0 \]
This can be factored as:
\[ (x - 1)^2 = 0 \]

**Hint:** What does this imply about \( x \)?

### Step 5: Solve for \( x \)
Setting the factor equal to zero gives:
\[ x - 1 = 0 \implies x = 1 \]
Thus, we have:
\[ \tan^2 \theta = 1 \]

**Hint:** Recall the values of \( \theta \) for which \( \tan \theta = 1 \).

### Step 6: Find the values of \( \theta \)
The solutions for \( \tan \theta = 1 \) are:
\[ \theta = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z} \]
This accounts for all angles where the tangent function equals 1.

**Hint:** Remember that the tangent function has a period of \( \pi \).

### Final Solution
The general solution for the equation \( \tan^2 \theta + \cot^2 \theta = 2 \) is:
\[ \theta = n\pi + \frac{\pi}{4}, \quad n \in \mathbb{Z} \]