If `theta` is a positive acute angle and `4cos^(2)theta-4costheta+1=0` , then the value of `tan(theta-15^(@))` is equal to

Options

A. 0
B. 1
C. `sqrt(3)`
D. `(1)/(sqrt(3))`

Answer

To solve the equation \( 4\cos^2\theta - 4\cos\theta + 1 = 0 \) and find the value of \( \tan(\theta - 15^\circ) \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 4\cos^2\theta - 4\cos\theta + 1 = 0 \]

### Step 2: Factor the quadratic equation
We can recognize that this is a quadratic equation in terms of \( \cos\theta \). We can rewrite it as:
\[ (2\cos\theta - 1)^2 = 0 \]
This is because \( 4\cos^2\theta - 4\cos\theta + 1 \) can be factored as \( (2\cos\theta - 1)(2\cos\theta - 1) \).

### Step 3: Solve for \( \cos\theta \)
Setting the factored equation to zero gives us:
\[ 2\cos\theta - 1 = 0 \]
Solving for \( \cos\theta \):
\[ 2\cos\theta = 1 \implies \cos\theta = \frac{1}{2} \]

### Step 4: Determine the angle \( \theta \)
Since \( \theta \) is a positive acute angle, we find:
\[ \theta = 60^\circ \]

### Step 5: Calculate \( \tan(\theta - 15^\circ) \)
Now we need to find:
\[ \tan(\theta - 15^\circ) = \tan(60^\circ - 15^\circ) = \tan(45^\circ) \]

### Step 6: Find the value of \( \tan(45^\circ) \)
We know that:
\[ \tan(45^\circ) = 1 \]

### Final Answer
Thus, the value of \( \tan(\theta - 15^\circ) \) is:
\[ \boxed{1} \]

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