Suppose`alpha ,beta , gamma in ` R are such that sin `alpha, sin beta , sin gamma ne 0`
and `Delta = |{:(sin^(2) alpha , sin alpha cos alpha , cos^(2) alpha),(sin^(2) beta , sin beta cos beta , cos^(2) beta),(sin^(2) gamma , sin gamma cos gamma , cos^(2) gamma):}|` then `Delta` cannot exceed

Options

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

Answer

To solve the problem, we need to evaluate the determinant \( \Delta \) given by:

\[ \Delta = \begin{vmatrix} \sin^2 \alpha & \sin \alpha \cos \alpha & \cos^2 \alpha \\ \sin^2 \beta & \sin \beta \cos \beta & \cos^2 \beta \\ \sin^2 \gamma & \sin \gamma \cos \gamma & \cos^2 \gamma \end{vmatrix} \]

### Step 1: Simplifying the Determinant
We can simplify the determinant by manipulating the columns. Notice that:

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

So, we can replace the first column with the sum of the first and third columns:

\[ \text{Column 1} = \text{Column 1} + \text{Column 3} \]

This gives us:

\[ \Delta = \begin{vmatrix} 1 & \sin \alpha \cos \alpha & 1 \\ 1 & \sin \beta \cos \beta & 1 \\ 1 & \sin \gamma \cos \gamma & 1 \end{vmatrix} \]

### Step 2: Factor out common terms
Next, we can factor out a 2 from the second column and a 2 from the third column:

\[ \Delta = \frac{1}{2} \begin{vmatrix} 1 & 2 \sin \alpha \cos \alpha & 2 \cos^2 \alpha \\ 1 & 2 \sin \beta \cos \beta & 2 \cos^2 \beta \\ 1 & 2 \sin \gamma \cos \gamma & 2 \cos^2 \gamma \end{vmatrix} \]

### Step 3: Further Simplification
Now we can express \( 2 \sin \theta \cos \theta \) as \( \sin(2\theta) \) and \( 2 \cos^2 \theta - 1 \) as \( \cos(2\theta) \):

\[ \Delta = \frac{1}{4} \begin{vmatrix} 1 & \sin(2\alpha) & \cos(2\alpha) \\ 1 & \sin(2\beta) & \cos(2\beta) \\ 1 & \sin(2\gamma) & \cos(2\gamma) \end{vmatrix} \]

### Step 4: Evaluating the Determinant
Now, we can evaluate the determinant using the properties of sine and cosine:

\[ \Delta = \frac{1}{4} \left( \sin(2\beta) \cos(2\gamma) - \sin(2\gamma) \cos(2\beta) + \sin(2\alpha) \cos(2\gamma) - \sin(2\gamma) \cos(2\alpha) + \sin(2\alpha) \cos(2\beta) - \sin(2\beta) \cos(2\alpha) \right) \]

### Step 5: Using Sine Difference Identity
This can be simplified further using the sine difference identity:

\[ \Delta = \frac{1}{4} \left( \sin(2\beta - 2\gamma) + \sin(2\alpha - 2\gamma) + \sin(2\alpha - 2\beta) \right) \]

### Step 6: Finding the Maximum Value
Since the sine function has a maximum value of 1, we can conclude:

\[ \Delta \leq \frac{3}{4} \]

Thus, the maximum value of \( \Delta \) cannot exceed 1.

### Conclusion
Therefore, we conclude that \( \Delta \) cannot exceed 1.