Question

If \[ \begin{vmatrix} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix} = \begin{vmatrix} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix} \] then the value of \( \cos^2 \alpha + \cos^3 \beta + \cos^2 \gamma \) is:

• A. \( \frac{1}{2} \)
• B. 1
• C. \( \frac{3}{2} \)
• D. \( \frac{1}{4} \)

Hint:

To solve determinant equations involving trigonometric functions, use cofactor expansion and simplify the expressions using trigonometric identities.

Answer 1

The Correct Option is B

Step 1: Analyze the Determinants.
We are given two determinants. First, let's compute the determinant on the left-hand side: \[ \text{Determinant 1} = \begin{vmatrix} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix} \] Expanding this determinant using cofactor expansion, we get a result that simplifies based on the trigonometric identities of \( \alpha, \beta, \gamma \).
Step 2: Compute the Right-Hand Side Determinant.
Similarly, compute the second determinant: \[ \text{Determinant 2} = \begin{vmatrix} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix} \] After solving this determinant, we observe that both determinants are equal under certain trigonometric relationships between \( \alpha, \beta, \gamma \).
Step 3: Relate the Determinants to the Expression.
After solving the system, we find that: \[ \cos^2 \alpha + \cos^3 \beta + \cos^2 \gamma = 1 \] Final Answer: \[ \boxed{1} \]