If tan `theta + (1)/(tan theta) = 2,` then
`tan^2 theta + (1)/(tan^2 theta)` is equal to

Options

A. 1
B. 2
C. 4
D. None of these

Answer

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

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

### Step 1: Square both sides
We square both sides of the equation to find an expression for \(\tan^2 \theta + \frac{1}{\tan^2 \theta}\).

\[ \left(\tan \theta + \frac{1}{\tan \theta}\right)^2 = 2^2 \]

### Step 2: Apply the square of a binomial identity
Using the identity \((a + b)^2 = a^2 + b^2 + 2ab\), we can expand the left-hand side:

\[ \tan^2 \theta + 2 \cdot \tan \theta \cdot \frac{1}{\tan \theta} + \frac{1}{\tan^2 \theta} = 4 \]

### Step 3: Simplify the equation
Since \(\tan \theta \cdot \frac{1}{\tan \theta} = 1\), we can simplify the equation:

\[ \tan^2 \theta + 2 + \frac{1}{\tan^2 \theta} = 4 \]

### Step 4: Isolate \(\tan^2 \theta + \frac{1}{\tan^2 \theta}\)
Now, we can isolate \(\tan^2 \theta + \frac{1}{\tan^2 \theta}\):

\[ \tan^2 \theta + \frac{1}{\tan^2 \theta} = 4 - 2 \]

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

### Final Answer
Thus, the value of \(\tan^2 \theta + \frac{1}{\tan^2 \theta}\) is:

\[ \boxed{2} \]

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