Question:
A value of \( \theta \) lying between \( 0 \) and \( \pi / 2 \) 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 value of \( \theta \) lying between \( 0 \) and \( \pi / 2 \) 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:
Show Hint
When a determinant contains cyclic or repeating trigonometric functions across rows or columns, look for column operations like \(C_1 + C_2\) to exploit the standard identity \(\sin^2\theta + \cos^2\theta = 1\). This instantly reduces variables to constants!
Updated On: May 20, 2026
- \( \frac{5\pi}{24} \)
- \( \frac{7\pi}{24} \)
- \( \frac{\pi}{8} \)
- \( \frac{3\pi}{8} \)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
Determinants can be simplified significantly before expansion by applying elementary row operations (\(R_i \rightarrow R_i + k R_j\)) or column operations (\(C_i \rightarrow C_i + k C_j\)). These operations preserve the value of the determinant.
Key trigonometric identity utilized:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
Step 1: Applying a column transformation to combine the trigonometric terms.
Let the given determinant be \(\Delta\). Apply the column operation \(C_1 \rightarrow C_1 + C_2\): \[ \Delta = \begin{vmatrix} (1 + \sin^2 \theta + \cos^2 \theta) & \cos^2 \theta & 4 \sin 4\theta (\sin^2 \theta + 1 + \cos^2 \theta) & 1 + \cos^2 \theta & 4 \sin 4\theta (\sin^2 \theta + \cos^2 \theta) & \cos^2 \theta & 1 + 4 \sin 4\theta \end{vmatrix} = 0 \] Using \(\sin^2 \theta + \cos^2 \theta = 1\), the first column simplifies to: \[ \begin{vmatrix} 2 & \cos^2 \theta & 4 \sin 4\theta 2 & 1 + \cos^2 \theta & 4 \sin 4\theta 1 & \cos^2 \theta & 1 + 4 \sin 4\theta \end{vmatrix} = 0 \]
Step 2: Applying row transformations to create maximum zeros in the determinant.
Perform the row operations \(R_1 \rightarrow R_1 - 2R_3\) and \(R_2 \rightarrow R_2 - 2R_3\): \[ \begin{vmatrix} 2 - 2(1) & \cos^2 \theta - 2\cos^2 \theta & 4 \sin 4\theta - 2(1 + 4 \sin 4\theta) 2 - 2(1) & 1 + \cos^2 \theta - 2\cos^2 \theta & 4 \sin 4\theta - 2(1 + 4 \sin 4\theta) 1 & \cos^2 \theta & 1 + 4 \sin 4\theta \end{vmatrix} = 0 \] Simplifying the terms: \[ \begin{vmatrix} 0 & -\cos^2 \theta & -2 - 4 \sin 4\theta 0 & 1 - \cos^2 \theta & -2 - 4 \sin 4\theta 1 & \cos^2 \theta & 1 + 4 \sin 4\theta \end{vmatrix} = 0 \]
Step 3: Expanding the determinant along the first column.
Expanding along \(C_1\): \[ 1 \cdot \begin{vmatrix} -\cos^2 \theta & -2 - 4 \sin 4\theta 1 - \cos^2 \theta & -2 - 4 \sin 4\theta \end{vmatrix} = 0 \] Taking out the common factor \(-(2 + 4 \sin 4\theta)\) from the second column: \[ -(2 + 4 \sin 4\theta) \begin{vmatrix} -\cos^2 \theta & 1 1 - \cos^2 \theta & 1 \end{vmatrix} = 0 \] Evaluating the \(2 \times 2\) determinant: \[ -\cos^2 \theta(1) - (1 - \cos^2 \theta)(1) = -\cos^2 \theta - 1 + \cos^2 \theta = -1 \] Substituting this back gives: \[ -(2 + 4 \sin 4\theta)(-1) = 0 \quad \Rightarrow \quad 2 + 4 \sin 4\theta = 0 \] \[ 4 \sin 4\theta = -2 \quad \Rightarrow \quad \sin 4\theta = -\frac{1}{2} \]
Step 4: Solving for \(\theta\) in the given interval \(0 < \theta < \pi/2\).
Given \(0 < \theta < \frac{\pi}{2}\), multiplying by 4 gives the range for \(4\theta\): \[ 0 < 4\theta < 2\pi \] In the interval \((0, 2\pi)\), \(\sin 4\theta = -\frac{1}{2}\) occurs in the 3rd and 4th quadrants: \[ 4\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \quad \Rightarrow \quad \theta = \frac{7\pi}{24} \] \[ 4\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \quad \Rightarrow \quad \theta = \frac{11\pi}{24} \] Comparing with the given options, \(\theta = \frac{7\pi}{24}\) is option (B).
Step 1: Applying a column transformation to combine the trigonometric terms.
Let the given determinant be \(\Delta\). Apply the column operation \(C_1 \rightarrow C_1 + C_2\): \[ \Delta = \begin{vmatrix} (1 + \sin^2 \theta + \cos^2 \theta) & \cos^2 \theta & 4 \sin 4\theta (\sin^2 \theta + 1 + \cos^2 \theta) & 1 + \cos^2 \theta & 4 \sin 4\theta (\sin^2 \theta + \cos^2 \theta) & \cos^2 \theta & 1 + 4 \sin 4\theta \end{vmatrix} = 0 \] Using \(\sin^2 \theta + \cos^2 \theta = 1\), the first column simplifies to: \[ \begin{vmatrix} 2 & \cos^2 \theta & 4 \sin 4\theta 2 & 1 + \cos^2 \theta & 4 \sin 4\theta 1 & \cos^2 \theta & 1 + 4 \sin 4\theta \end{vmatrix} = 0 \]
Step 2: Applying row transformations to create maximum zeros in the determinant.
Perform the row operations \(R_1 \rightarrow R_1 - 2R_3\) and \(R_2 \rightarrow R_2 - 2R_3\): \[ \begin{vmatrix} 2 - 2(1) & \cos^2 \theta - 2\cos^2 \theta & 4 \sin 4\theta - 2(1 + 4 \sin 4\theta) 2 - 2(1) & 1 + \cos^2 \theta - 2\cos^2 \theta & 4 \sin 4\theta - 2(1 + 4 \sin 4\theta) 1 & \cos^2 \theta & 1 + 4 \sin 4\theta \end{vmatrix} = 0 \] Simplifying the terms: \[ \begin{vmatrix} 0 & -\cos^2 \theta & -2 - 4 \sin 4\theta 0 & 1 - \cos^2 \theta & -2 - 4 \sin 4\theta 1 & \cos^2 \theta & 1 + 4 \sin 4\theta \end{vmatrix} = 0 \]
Step 3: Expanding the determinant along the first column.
Expanding along \(C_1\): \[ 1 \cdot \begin{vmatrix} -\cos^2 \theta & -2 - 4 \sin 4\theta 1 - \cos^2 \theta & -2 - 4 \sin 4\theta \end{vmatrix} = 0 \] Taking out the common factor \(-(2 + 4 \sin 4\theta)\) from the second column: \[ -(2 + 4 \sin 4\theta) \begin{vmatrix} -\cos^2 \theta & 1 1 - \cos^2 \theta & 1 \end{vmatrix} = 0 \] Evaluating the \(2 \times 2\) determinant: \[ -\cos^2 \theta(1) - (1 - \cos^2 \theta)(1) = -\cos^2 \theta - 1 + \cos^2 \theta = -1 \] Substituting this back gives: \[ -(2 + 4 \sin 4\theta)(-1) = 0 \quad \Rightarrow \quad 2 + 4 \sin 4\theta = 0 \] \[ 4 \sin 4\theta = -2 \quad \Rightarrow \quad \sin 4\theta = -\frac{1}{2} \]
Step 4: Solving for \(\theta\) in the given interval \(0 < \theta < \pi/2\).
Given \(0 < \theta < \frac{\pi}{2}\), multiplying by 4 gives the range for \(4\theta\): \[ 0 < 4\theta < 2\pi \] In the interval \((0, 2\pi)\), \(\sin 4\theta = -\frac{1}{2}\) occurs in the 3rd and 4th quadrants: \[ 4\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \quad \Rightarrow \quad \theta = \frac{7\pi}{24} \] \[ 4\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \quad \Rightarrow \quad \theta = \frac{11\pi}{24} \] Comparing with the given options, \(\theta = \frac{7\pi}{24}\) is option (B).
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