Suppose `alpha, beta, gamma, theta epsilon R` and
`A(alpha, beta,gamma, theta)=|(cos (alpha + theta), sin (alpha+theta),1),(cos(beta+theta),sin(beta+theta),1),(cos (gamma+theta),sin(gamma+theta),1)|`
Numerical value of `A(-(pi)/2,0,(pi)/2,(2pi)/13)` is

Options

A. 0
B. `-1`
C. 2
D. None of these

Answer

To find the numerical value of \( A(-\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{2\pi}{13}) \), we start with the determinant defined in the question:

\[ A(\alpha, \beta, \gamma, \theta) = \begin{vmatrix} \cos(\alpha + \theta) & \sin(\alpha + \theta) & 1 \\ \cos(\beta + \theta) & \sin(\beta + \theta) & 1 \\ \cos(\gamma + \theta) & \sin(\gamma + \theta) & 1 \end{vmatrix} \]

### Step 1: Substitute the values of \(\alpha\), \(\beta\), \(\gamma\), and \(\theta\)

We need to substitute \(\alpha = -\frac{\pi}{2}\), \(\beta = 0\), \(\gamma = \frac{\pi}{2}\), and \(\theta = \frac{2\pi}{13}\):

\[ A\left(-\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{2\pi}{13}\right) = \begin{vmatrix} \cos\left(-\frac{\pi}{2} + \frac{2\pi}{13}\right) & \sin\left(-\frac{\pi}{2} + \frac{2\pi}{13}\right) & 1 \\ \cos\left(0 + \frac{2\pi}{13}\right) & \sin\left(0 + \frac{2\pi}{13}\right) & 1 \\ \cos\left(\frac{\pi}{2} + \frac{2\pi}{13}\right) & \sin\left(\frac{\pi}{2} + \frac{2\pi}{13}\right) & 1 \end{vmatrix} \]

### Step 2: Simplify the trigonometric functions

Now we calculate the trigonometric values:

1. For \(\alpha = -\frac{\pi}{2}\):
- \(\cos\left(-\frac{\pi}{2} + \frac{2\pi}{13}\right) = \sin\left(\frac{2\pi}{13}\right)\)
- \(\sin\left(-\frac{\pi}{2} + \frac{2\pi}{13}\right) = -\cos\left(\frac{2\pi}{13}\right)\)

2. For \(\beta = 0\):
- \(\cos\left(0 + \frac{2\pi}{13}\right) = \cos\left(\frac{2\pi}{13}\right)\)
- \(\sin\left(0 + \frac{2\pi}{13}\right) = \sin\left(\frac{2\pi}{13}\right)\)

3. For \(\gamma = \frac{\pi}{2}\):
- \(\cos\left(\frac{\pi}{2} + \frac{2\pi}{13}\right) = -\sin\left(\frac{2\pi}{13}\right)\)
- \(\sin\left(\frac{\pi}{2} + \frac{2\pi}{13}\right) = \cos\left(\frac{2\pi}{13}\right)\)

Thus, the determinant becomes:

\[ A\left(-\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{2\pi}{13}\right) = \begin{vmatrix} \sin\left(\frac{2\pi}{13}\right) & -\cos\left(\frac{2\pi}{13}\right) & 1 \\ \cos\left(\frac{2\pi}{13}\right) & \sin\left(\frac{2\pi}{13}\right) & 1 \\ -\sin\left(\frac{2\pi}{13}\right) & \cos\left(\frac{2\pi}{13}\right) & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant

Now we can calculate the determinant using the formula for a \(3 \times 3\) matrix:

\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Substituting the values:

\[ \text{Det} = \sin\left(\frac{2\pi}{13}\right) \left(\sin\left(\frac{2\pi}{13}\right) \cdot 1 - \cos\left(\frac{2\pi}{13}\right) \cdot 1\right) - (-\cos\left(\frac{2\pi}{13}\right)) \left(\cos\left(\frac{2\pi}{13}\right) \cdot 1 - (-\sin\left(\frac{2\pi}{13}\right) \cdot 1)\right) + 1 \left(\cos\left(\frac{2\pi}{13}\right) \cdot \cos\left(\frac{2\pi}{13}\right) - (-\sin\left(\frac{2\pi}{13}\right) \cdot \sin\left(\frac{2\pi}{13}\right))\right) \]

### Step 4: Simplify the expression

After simplifying the determinant, we find:

\[ = \sin^2\left(\frac{2\pi}{13}\right) + \cos^2\left(\frac{2\pi}{13}\right) = 1 \]

### Final Result

Thus, the numerical value of \( A\left(-\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{2\pi}{13}\right) \) is:

\[ \boxed{1} \]