Question

The following determinant is equal to:

\[ \begin{vmatrix} \sin^2 x & \cos^2 x & 1 \\ \cos^2 x & \sin^2 x & 1 \\ -10 & 12 & 2 \end{vmatrix} \]

• A. 0
• B. \(12\cos^2 x - 10\sin^2 x\)
• C. \(12\cos^2 x - 10\sin^2 x - 2\)
• D. \(10\sin 2x\)

Hint:

To simplify determinant calculations: Expand along a row/column with zeros or simple entries. Recognize patterns like symmetric determinants or factorable expressions. Use determinant properties to simplify before expansion.

Answer 1

The Correct Option is A

We are given the determinant:

\[ D = \begin{vmatrix} \sin^2 x & \cos^2 x & 1 \\ \cos^2 x & \sin^2 x & 1 \\ -10 & 12 & 2 \end{vmatrix} \]

We evaluate this determinant using cofactor expansion along the first row:

\[ D = \sin^2 x \begin{vmatrix} \sin^2 x & 1 \\ 12 & 2 \end{vmatrix} - \cos^2 x \begin{vmatrix} \cos^2 x & 1 \\ -10 & 2 \end{vmatrix} + 1 \begin{vmatrix} \cos^2 x & \sin^2 x \\ -10 & 12 \end{vmatrix} \]

Step 1: Compute the \(2 \times 2\) Determinants

Using the determinant formula for a \(2 \times 2\) matrix:

\[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \]

First determinant:

\[ \begin{vmatrix} \sin^2 x & 1 \\ 12 & 2 \end{vmatrix} = (\sin^2 x \cdot 2) - (1 \cdot 12) = 2\sin^2 x - 12 \]

Second determinant:

\[ \begin{vmatrix} \cos^2 x & 1 \\ -10 & 2 \end{vmatrix} = (\cos^2 x \cdot 2) - (1 \cdot (-10)) = 2\cos^2 x + 10 \]

Third determinant:

\[ \begin{vmatrix} \cos^2 x & \sin^2 x \\ -10 & 12 \end{vmatrix} = (\cos^2 x \cdot 12) - (\sin^2 x \cdot (-10)) = 12\cos^2 x + 10\sin^2 x \]

Step 2: Compute \(D\)

\[ D = \sin^2 x (2\sin^2 x - 12) - \cos^2 x (2\cos^2 x + 10) + (12\cos^2 x + 10\sin^2 x) \]

Expanding each term:

\[ D = 2\sin^4 x - 12\sin^2 x - 2\cos^4 x - 10\cos^2 x + 12\cos^2 x + 10\sin^2 x \]

Rearrange terms:

\[ D = 2\sin^4 x - 2\cos^4 x - 12\sin^2 x - 10\cos^2 x + 12\cos^2 x + 10\sin^2 x \]

Factor:

\[ D = 2(\sin^4 x - \cos^4 x) - (12\sin^2 x - 10\sin^2 x) + (12\cos^2 x - 10\cos^2 x) \]

Using the identity \( \sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) = (\sin^2 x - \cos^2 x)(1) \), we get:

\[ D = 2(\sin^2 x - \cos^2 x) - 2\sin^2 x + 2\cos^2 x \]

Since \( 2\cos^2 x - 2\cos^2 x = 0 \) and \( 2\sin^2 x - 2\sin^2 x = 0 \), we obtain:

\[ D = 0 \]

Final Answer:

\[ \boxed{0} \]