To solve the given problem, we need to find the values of the acute angles \( \alpha \) and \( \beta \) using the condition that \( \alpha + \beta \) and \( \alpha - \beta \) satisfy the equation:
\(\tan^2 \theta - 4\tan \theta + 1 = 0\)
Firstly, let's solve the quadratic equation \(\tan^2 \theta - 4\tan \theta + 1 = 0\):
The equation is of the form \(ax^2 + bx + c = 0\), where \( a = 1 \), \( b = -4 \), and \( c = 1 \).
The quadratic formula is given by:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Substitute the values of \( a \), \( b \), and \( c \):
\(x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}\)
This simplifies to:
\(x = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2}\)
\(x = 2 \pm \sqrt{3}\)
Thus, the solutions to the quadratic equation are:
\(\tan \theta = 2 + \sqrt{3}\) and \(\tan \theta = 2 - \sqrt{3}\)
Now, we know the values of \(\tan(\alpha + \beta)\) and \(\tan(\alpha - \beta)\) must satisfy these values. We need to check which angles \(\alpha\) and \(\beta\) will satisfy these conditions.
We use the known tangent values for certain special angles:
\(\tan 75^\circ = 2 + \sqrt{3}\)
\(\tan 15^\circ = 2 - \sqrt{3}\)
This means \(\alpha + \beta = 75^\circ\) and \(\alpha - \beta = 15^\circ\).
We can find \(\alpha\) and \(\beta\) by solving the system of equations:
Adding these two equations, we get:
\(2\alpha = 90^\circ \implies \alpha = 45^\circ\)
Substituting \(\alpha = 45^\circ\) into the first equation:
\(45^\circ + \beta = 75^\circ \implies \beta = 30^\circ\)
Therefore, the values of \(\alpha\) and \(\beta\) that satisfy the given conditions are \(45^\circ\) and \(30^\circ\) respectively.
The correct answer is: \(45^\circ, 30^\circ\)