If ` tan theta + cot theta =2, " then " tan^(2) theta + cot^(2) theta` is equal to

Options

A. 4
B. 3
C. 2
D. none of these

Explanation

To solve the problem, we start with the given equation:

\[ \tan \theta + \cot \theta = 2 \]

We need to find the value of:

\[ \tan^2 \theta + \cot^2 \theta \]

### Step 1: Rewrite cotangent in terms of tangent

We know that:

\[ \cot \theta = \frac{1}{\tan \theta} \]

Substituting this into the equation gives us:

\[ \tan \theta + \frac{1}{\tan \theta} = 2 \]

### Step 2: Multiply through by \(\tan \theta\)

To eliminate the fraction, we multiply both sides by \(\tan \theta\):

\[ \tan^2 \theta + 1 = 2 \tan \theta \]

### Step 3: Rearrange the equation

Rearranging the equation gives us:

\[ \tan^2 \theta - 2 \tan \theta + 1 = 0 \]

### Step 4: Factor the quadratic equation

This can be factored as:

\[ (\tan \theta - 1)^2 = 0 \]

### Step 5: Solve for \(\tan \theta\)

Setting the factor equal to zero gives us:

\[ \tan \theta - 1 = 0 \implies \tan \theta = 1 \]

### Step 6: Find \(\cot \theta\)

Using the value of \(\tan \theta\):

\[ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{1} = 1 \]

### Step 7: Calculate \(\tan^2 \theta + \cot^2 \theta\)

Now we can find:

\[ \tan^2 \theta + \cot^2 \theta = 1^2 + 1^2 = 1 + 1 = 2 \]

### Final Answer

Thus, the value of \(\tan^2 \theta + \cot^2 \theta\) is:

\[ \boxed{2} \]