Question

A value of \( \theta \) lying between \( 0 \) and \( \dfrac{\pi}{2} \) and satisfying \[ \begin{vmatrix} 1 + \sin^2 \theta & \cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4\sin 4\theta \end{vmatrix} = 0 \] is:

• A. \( \dfrac{5\pi}{24} \)
• B. \( \dfrac{7\pi}{24} \)
• C. \( \dfrac{\pi}{8} \)
• D. \( \dfrac{3\pi}{8} \)

Hint:

Use row operations to simplify determinants when the structure allows easy row elimination.

Answer 1

The Correct Option is B

Step 1: Let us denote the determinant as \( D \). Given: \[ D = \begin{vmatrix} 1 + \sin^2 \theta & \cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4\sin 4\theta \end{vmatrix} = 0 \]
Step 2: Perform row operation \( R_1 \rightarrow R_1 - R_2 \): \[ R_1 \rightarrow (1 + \sin^2 \theta - \sin^2 \theta, \cos^2 \theta - (1 + \cos^2 \theta), 4\sin 4\theta - 4\sin 4\theta) \] \[ \Rightarrow R_1 = (1, -1, 0) \]
Step 3: Perform row operation \( R_2 \rightarrow R_2 - R_3 \): \[ R_2 = (\sin^2 \theta - \sin^2 \theta, 1 + \cos^2 \theta - \cos^2 \theta, 4\sin 4\theta - (1 + 4\sin 4\theta)) = (0, 1, -1) \]
Step 4: So the matrix becomes: \[ D = \begin{vmatrix} 1 & -1 & 0
0 & 1 & -1
\sin^2 \theta & \cos^2 \theta & 1 + 4\sin 4\theta \end{vmatrix} \]
Step 5: Expand the determinant: \[ D = 1 . \begin{vmatrix} 1 & -1 \\ \cos^2 \theta & 1 + 4\sin 4\theta \end{vmatrix} + 1 . \begin{vmatrix} 0 & -1 \\ \sin^2 \theta & 1 + 4\sin 4\theta \end{vmatrix} \]
Step 6: Simplify: \[ = 1 . (1 . (1 + 4\sin 4\theta) + \cos^2 \theta) + 1 . (0 . (1 + 4\sin 4\theta) + \sin^2 \theta) \Rightarrow D = 1 + 4\sin 4\theta + \cos^2 \theta + \sin^2 \theta \] \[ = 1 + 4\sin 4\theta + 1 = 2 + 4\sin 4\theta \Rightarrow 2 + 4\sin 4\theta = 0 \Rightarrow \sin 4\theta = -\dfrac{1}{2} \]
Step 7: Find \( \theta \) such that \( 4\theta = 7\pi/6 \Rightarrow \theta = \dfrac{7\pi}{24} \) \[ \boxed{\theta = \dfrac{7\pi}{24}} \]