If `theta` is an acute angle and `tan^(2)theta+(1)/(tan^(2)theta)=2` , then the value of `theta` is :

Options

A. `60^(@)`
B. `45^(@)`
C. `15^(@)`
D. `30^(@)`

Answer

To solve the equation \( \tan^2 \theta + \frac{1}{\tan^2 \theta} = 2 \), we will follow these steps:

### Step 1: Set up the equation
We start with the given equation:
\[ \tan^2 \theta + \frac{1}{\tan^2 \theta} = 2 \]

### Step 2: Let \( x = \tan^2 \theta \)
We can substitute \( x \) for \( \tan^2 \theta \):
\[ x + \frac{1}{x} = 2 \]

### 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
This can be factored as:
\[ (x - 1)^2 = 0 \]

### Step 6: Solve for \( x \)
Setting the factor equal to zero gives:
\[ x - 1 = 0 \implies x = 1 \]

### Step 7: Substitute back for \( \tan^2 \theta \)
Since \( x = \tan^2 \theta \), we have:
\[ \tan^2 \theta = 1 \]

### Step 8: Solve for \( \theta \)
Taking the square root of both sides, we find:
\[ \tan \theta = 1 \]
Since \( \theta \) is an acute angle, we take the positive root:
\[ \theta = 45^\circ \]

### Final Answer
Thus, the value of \( \theta \) is:
\[ \theta = 45^\circ \]