If `alpha` and `beta` are acute such that `alpha+beta` and `alpha-beta` satisfy the equation `tan^(2)theta-4tan theta+1=0`, then `(alpha, beta`) =

Options

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

Explanation

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

### Step 1: Solve the quadratic equation
The given equation is:
\[ \tan^2 \theta - 4 \tan \theta + 1 = 0 \]
This is a quadratic equation in terms of \(\tan \theta\). We can use the quadratic formula:
\[ \tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -4\), and \(c = 1\).

### Step 2: Calculate the discriminant
Calculating the discriminant:
\[ b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 1 = 16 - 4 = 12 \]

### Step 3: Substitute values into the quadratic formula
Now substituting the values into the quadratic formula:
\[ \tan \theta = \frac{4 \pm \sqrt{12}}{2} \]
\[ \tan \theta = \frac{4 \pm 2\sqrt{3}}{2} \]
\[ \tan \theta = 2 \pm \sqrt{3} \]

### Step 4: Identify the angles
Thus, we have two values:
1. \(\tan(\alpha + \beta) = 2 + \sqrt{3}\)
2. \(\tan(\alpha - \beta) = 2 - \sqrt{3}\)

### Step 5: Find the angles corresponding to the tangent values
We recognize that:
\[ \tan(75^\circ) = 2 + \sqrt{3} \]
and
\[ \tan(15^\circ) = 2 - \sqrt{3} \]

### Step 6: Set up equations for \(\alpha\) and \(\beta\)
From the above, we can set up the equations:
1. \(\alpha + \beta = 75^\circ\) (Equation 1)
2. \(\alpha - \beta = 15^\circ\) (Equation 2)

### Step 7: Solve the system of equations
Now, we can add Equation 1 and Equation 2:
\[ (\alpha + \beta) + (\alpha - \beta) = 75^\circ + 15^\circ \]
\[ 2\alpha = 90^\circ \]
\[ \alpha = 45^\circ \]

Next, substitute \(\alpha\) back into Equation 1 to find \(\beta\):
\[ 45^\circ + \beta = 75^\circ \]
\[ \beta = 75^\circ - 45^\circ = 30^\circ \]

### Final Answer
Thus, the values of \(\alpha\) and \(\beta\) are:
\[ (\alpha, \beta) = (45^\circ, 30^\circ) \]