Question:
If \(\Delta_1 = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}\) and \(\Delta_2 = \begin{vmatrix} x & \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x & 1 \\ \cos 2\theta & 1 & x \end{vmatrix}\), \(x \ne 0\); then for all \(\theta \in \left(0, \frac{\pi}{2}\right)\) :
If \(\Delta_1 = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}\) and \(\Delta_2 = \begin{vmatrix} x & \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x & 1 \\ \cos 2\theta & 1 & x \end{vmatrix}\), \(x \ne 0\); then for all \(\theta \in \left(0, \frac{\pi}{2}\right)\) :
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Expand \(\Delta_1\) along the first row; the \(\theta\)-terms cancel and \(\Delta_1=-x^3\), and the same computation gives \(\Delta_2=-x^3\).
Updated On: Nov 5, 2026
- \(\Delta_1 - \Delta_2 = x(\cos 2\theta - \cos 4\theta)\)
- \(\Delta_1 + \Delta_2 = -2x^3\)
- \(\Delta_1 + \Delta_2 = -2(x^3 + x - 1)\)
- \(\Delta_1 - \Delta_2 = -2x^3\)
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The Correct Option is B
Solution and Explanation
Expand \(\Delta_1\) along row 1:
\(\Delta_1 = x\left[(-x)(x) - 1\cdot1\right] - \sin\theta\left[(-\sin\theta)(x) - 1\cdot\cos\theta\right] + \cos\theta\left[(-\sin\theta)(1) - (-x)\cos\theta\right]\)
\(= -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \sin\theta\cos\theta + x\cos^2\theta\)
\(= -x^3 - x + x(\sin^2\theta+\cos^2\theta) = -x^3\).
\(\Delta_2\) has the same form with \(\theta\) replaced by \(2\theta\), so \(\Delta_2 = -x^3\) too.
Therefore \(\Delta_1 - \Delta_2 = 0\) and \(\Delta_1 + \Delta_2 = -2x^3\).
Answer: (2) \(\Delta_1 + \Delta_2 = -2x^3\)
\(\Delta_1 = x\left[(-x)(x) - 1\cdot1\right] - \sin\theta\left[(-\sin\theta)(x) - 1\cdot\cos\theta\right] + \cos\theta\left[(-\sin\theta)(1) - (-x)\cos\theta\right]\)
\(= -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \sin\theta\cos\theta + x\cos^2\theta\)
\(= -x^3 - x + x(\sin^2\theta+\cos^2\theta) = -x^3\).
\(\Delta_2\) has the same form with \(\theta\) replaced by \(2\theta\), so \(\Delta_2 = -x^3\) too.
Therefore \(\Delta_1 - \Delta_2 = 0\) and \(\Delta_1 + \Delta_2 = -2x^3\).
Answer: (2) \(\Delta_1 + \Delta_2 = -2x^3\)
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