if `abs({:(sinalpha, cosbeta), (cosalpha, sinbeta):})=1/2`, where `alpha, beta` are acute angles, then find the value of `alpha+beta`. a) `pi/3` b) `(2pi)/3``c) `pi/2` d) `pi/4`

Options

A. `pi/3`
B. `(2pi)/3`
C. `pi/2`
D. `pi/4`

Answer

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

\[ \text{abs}\begin{vmatrix} \sin \alpha & \cos \beta \\ \cos \alpha & \sin \beta \end{vmatrix} = \frac{1}{2} \]

### Step 1: Calculate the determinant

The determinant of a 2x2 matrix is calculated as follows:

\[ \text{det} = (\sin \alpha)(\sin \beta) - (\cos \alpha)(\cos \beta) \]

So we can write:

\[ |\sin \alpha \sin \beta - \cos \alpha \cos \beta| = \frac{1}{2} \]

### Step 2: Simplify the expression

Using the identity for cosine, we know:

\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]

Thus, we can rewrite our determinant:

\[ \sin \alpha \sin \beta - \cos \alpha \cos \beta = -(\cos \alpha \cos \beta - \sin \alpha \sin \beta) = -\cos(\alpha + \beta) \]

This gives us:

\[ |-\cos(\alpha + \beta)| = \frac{1}{2} \]

### Step 3: Remove the absolute value

Since \(\alpha\) and \(\beta\) are acute angles, \(\alpha + \beta\) will also be an acute angle or less than \(\pi\). Therefore, we can remove the absolute value:

\[ -\cos(\alpha + \beta) = \frac{1}{2} \quad \text{or} \quad \cos(\alpha + \beta) = -\frac{1}{2} \]

### Step 4: Solve for \(\alpha + \beta\)

The cosine function equals \(-\frac{1}{2}\) at specific angles. The angle that satisfies this in the range of \(0\) to \(\pi\) is:

\[ \alpha + \beta = \frac{2\pi}{3} \]

### Final Answer

Thus, the value of \(\alpha + \beta\) is:

\[ \alpha + \beta = \frac{2\pi}{3} \]