If `tantheta + cot theta = 2` , find the value of `tan^2 theta + cot^2 theta`

Answer

To solve the problem, we need to find the value of \( \tan^2 \theta + \cot^2 \theta \) given that \( \tan \theta + \cot \theta = 2 \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \tan \theta + \cot \theta = 2 \]

2. **Square both sides of the equation:**
\[ (\tan \theta + \cot \theta)^2 = 2^2 \]
This simplifies to:
\[ \tan^2 \theta + 2 \tan \theta \cot \theta + \cot^2 \theta = 4 \]

3. **Use the identity \( \cot \theta = \frac{1}{\tan \theta} \):**
Substitute \( \cot \theta \) in the equation:
\[ \tan^2 \theta + 2 \tan \theta \left(\frac{1}{\tan \theta}\right) + \left(\frac{1}{\tan \theta}\right)^2 = 4 \]
This simplifies to:
\[ \tan^2 \theta + 2 + \frac{1}{\tan^2 \theta} = 4 \]

4. **Rearrange the equation:**
\[ \tan^2 \theta + \frac{1}{\tan^2 \theta} = 4 - 2 \]
This simplifies to:
\[ \tan^2 \theta + \frac{1}{\tan^2 \theta} = 2 \]

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

6. **Multiply through by \( x \) to eliminate the fraction:**
\[ x^2 + 1 = 2x \]

7. **Rearrange to form a quadratic equation:**
\[ x^2 - 2x + 1 = 0 \]

8. **Factor the quadratic equation:**
\[ (x - 1)^2 = 0 \]

9. **Solve for \( x \):**
\[ x - 1 = 0 \implies x = 1 \]

10. **Substitute back to find \( \tan^2 \theta + \cot^2 \theta \):**
Since \( x = \tan^2 \theta \), we have:
\[ \tan^2 \theta = 1 \quad \text{and} \quad \cot^2 \theta = \frac{1}{\tan^2 \theta} = 1 \]
Therefore:
\[ \tan^2 \theta + \cot^2 \theta = 1 + 1 = 2 \]

### Final Answer:
\[ \tan^2 \theta + \cot^2 \theta = 2 \]