Question:
Let \(\theta \neq \dfrac{(2n+1)\pi}{2}\), for \(n\in\mathbb{Z}\). Consider the system of equations
\[
x\cos\theta+y\sec\theta=0
\]
\[
x\sin\theta+y\tan\theta=0
\]
This system does not have a unique solution if and only if \(\theta\) belongs to:
Let \(\theta \neq \dfrac{(2n+1)\pi}{2}\), for \(n\in\mathbb{Z}\). Consider the system of equations
\[
x\cos\theta+y\sec\theta=0
\]
\[
x\sin\theta+y\tan\theta=0
\]
This system does not have a unique solution if and only if \(\theta\) belongs to:
Show Hint
For homogeneous systems, always check the determinant. If determinant is non-zero, the only solution is the trivial solution. If determinant is zero, infinitely many or non-unique solutions become possible.
Updated On: Jun 11, 2026
- \(\{n\pi:n\in\mathbb{Z}\}\)
- \(\{(2n+1)\pi:n\in\mathbb{Z}\}\)
- \(\{2n\pi:n\in\mathbb{Z}\}\)
- \(\{(4n+1)\pi:n\in\mathbb{Z}\}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
A homogeneous system of linear equations has a unique solution if and only if the determinant of the coefficient matrix is non-zero.
Therefore, the system will not have a unique solution when the determinant becomes zero.
Step 1: Write the coefficient matrix. The given equations are: \[ x\cos\theta+y\sec\theta=0 \] \[ x\sin\theta+y\tan\theta=0 \] Coefficient matrix: \[ A= \begin{bmatrix} \cos\theta & \sec\theta\\ \sin\theta & \tan\theta \end{bmatrix} \]
Step 2: Compute the determinant. \[ |A| = \cos\theta\tan\theta - \sin\theta\sec\theta \] Substituting: \[ \tan\theta=\frac{\sin\theta}{\cos\theta} \] \[ \sec\theta=\frac{1}{\cos\theta} \] Hence, \[ |A| = \cos\theta \left( \frac{\sin\theta}{\cos\theta} \right) - \sin\theta \left( \frac{1}{\cos\theta} \right) \] \[ = \sin\theta - \frac{\sin\theta}{\cos\theta} \] \[ = \frac{\sin\theta(\cos\theta-1)} {\cos\theta} \]
Step 3: Set determinant equal to zero. \[ \frac{\sin\theta(\cos\theta-1)} {\cos\theta}=0 \] Since \(\cos\theta\neq0\), \[ \sin\theta(\cos\theta-1)=0 \] Therefore, \[ \sin\theta=0 \] or \[ \cos\theta=1 \]
Step 4: Solve the conditions. From \[ \sin\theta=0 \] we obtain \[ \theta=n\pi \] The condition \(\cos\theta=1\) gives \[ \theta=2n\pi \] which is already included in \[ \theta=n\pi \] Therefore, \[ \boxed{\theta=n\pi} \] Hence the system does not possess a unique solution when \[ \theta\in\{n\pi:n\in\mathbb{Z}\} \] Therefore option (A) is correct.
Step 1: Write the coefficient matrix. The given equations are: \[ x\cos\theta+y\sec\theta=0 \] \[ x\sin\theta+y\tan\theta=0 \] Coefficient matrix: \[ A= \begin{bmatrix} \cos\theta & \sec\theta\\ \sin\theta & \tan\theta \end{bmatrix} \]
Step 2: Compute the determinant. \[ |A| = \cos\theta\tan\theta - \sin\theta\sec\theta \] Substituting: \[ \tan\theta=\frac{\sin\theta}{\cos\theta} \] \[ \sec\theta=\frac{1}{\cos\theta} \] Hence, \[ |A| = \cos\theta \left( \frac{\sin\theta}{\cos\theta} \right) - \sin\theta \left( \frac{1}{\cos\theta} \right) \] \[ = \sin\theta - \frac{\sin\theta}{\cos\theta} \] \[ = \frac{\sin\theta(\cos\theta-1)} {\cos\theta} \]
Step 3: Set determinant equal to zero. \[ \frac{\sin\theta(\cos\theta-1)} {\cos\theta}=0 \] Since \(\cos\theta\neq0\), \[ \sin\theta(\cos\theta-1)=0 \] Therefore, \[ \sin\theta=0 \] or \[ \cos\theta=1 \]
Step 4: Solve the conditions. From \[ \sin\theta=0 \] we obtain \[ \theta=n\pi \] The condition \(\cos\theta=1\) gives \[ \theta=2n\pi \] which is already included in \[ \theta=n\pi \] Therefore, \[ \boxed{\theta=n\pi} \] Hence the system does not possess a unique solution when \[ \theta\in\{n\pi:n\in\mathbb{Z}\} \] Therefore option (A) is correct.
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