Consider the following statements in respect of the determinant `|("cos"^(2)(alpha)/(2),"sin"^(2)(alpha)/(2)),("sin"^(2) (beta)/(2),"cos"^(2) (beta)/(2))|` where `alpha, beta` are complementary angles.
I. The value of the determinant is `(1)/(sqrt(2)) "cos" ((alpha - beta)/(2))`.
II. The maximum value of the determinant is `(1)/(sqrt(2))`.
Which of the above statement(s) is/are correct ?

Options

A. A) Only I
B. B) Only II
C. C) Both I and II
D. D) Neither I nor II

Answer

To solve the given determinant problem, we need to evaluate the determinant:

\[ D = \begin{vmatrix} \frac{\cos^2(\alpha)}{2} & \frac{\sin^2(\alpha)}{2} \\ \frac{\sin^2(\beta)}{2} & \frac{\cos^2(\beta)}{2} \end{vmatrix} \]

where \(\alpha\) and \(\beta\) are complementary angles, meaning \(\alpha + \beta = 90^\circ\).

### Step 1: Calculate the determinant
Using the formula for the determinant of a 2x2 matrix:

\[ D = ad - bc \]

where \(a = \frac{\cos^2(\alpha)}{2}\), \(b = \frac{\sin^2(\alpha)}{2}\), \(c = \frac{\sin^2(\beta)}{2}\), and \(d = \frac{\cos^2(\beta)}{2}\).

Substituting the values, we get:

\[ D = \left(\frac{\cos^2(\alpha)}{2}\right) \left(\frac{\cos^2(\beta)}{2}\right) - \left(\frac{\sin^2(\alpha)}{2}\right) \left(\frac{\sin^2(\beta)}{2}\right) \]

This simplifies to:

\[ D = \frac{1}{4} \left(\cos^2(\alpha) \cos^2(\beta) - \sin^2(\alpha) \sin^2(\beta)\right) \]

### Step 2: Apply the identity for cosine
Using the identity \( \cos^2(A) - \sin^2(B) = \cos(A + B) \) for complementary angles, we can rewrite:

\[ D = \frac{1}{4} \left(\cos^2(\alpha) \cos^2(\beta) - \sin^2(\alpha) \sin^2(\beta)\right) = \frac{1}{4} \cdot \cos\left(\alpha + \beta\right) = \frac{1}{4} \cdot \cos(90^\circ) = 0 \]

However, we need to express this in terms of half angles. Using the half-angle formulas:

\[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}, \quad \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \]

We can express the determinant in terms of \(\alpha\) and \(\beta\).

### Step 3: Substitute complementary angles
Since \(\beta = 90^\circ - \alpha\), we have:

\[ \cos^2(\beta) = \sin^2(\alpha), \quad \sin^2(\beta) = \cos^2(\alpha) \]

Thus, we can rewrite the determinant:

\[ D = \frac{1}{4} \left(\cos^2(\alpha) \sin^2(\alpha) - \sin^2(\alpha) \cos^2(\alpha)\right) = \frac{1}{4} \left(\cos^2(\alpha) - \sin^2(\alpha)\right) \]

### Step 4: Final expression
Using the cosine of the difference of angles:

\[ D = \frac{1}{4} \cos(\alpha - \beta) \]

Since \(\beta = 90^\circ - \alpha\), we have:

\[ D = \frac{1}{4} \cos(\alpha - (90^\circ - \alpha)) = \frac{1}{4} \cos(2\alpha - 90^\circ) = \frac{1}{4} \sin(2\alpha) \]

### Step 5: Maximum value
The maximum value of \(\sin(2\alpha)\) is \(1\), thus:

\[ \text{Maximum value of } D = \frac{1}{4} \cdot 1 = \frac{1}{4} \]

### Conclusion
1. The value of the determinant is \(\frac{1}{4} \cos(\alpha - \beta)\) which is not equal to \(\frac{1}{\sqrt{2}} \cos\left(\frac{\alpha - \beta}{2}\right)\) as stated in statement I.
2. The maximum value of the determinant is \(\frac{1}{4}\), not \(\frac{1}{\sqrt{2}}\) as stated in statement II.

Thus, both statements are incorrect.