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:

• A. \(\{n\pi:n\in\mathbb{Z}\}\)
• B. \(\{(2n+1)\pi:n\in\mathbb{Z}\}\)
• C. \(\{2n\pi:n\in\mathbb{Z}\}\)
• D. \(\{(4n+1)\pi:n\in\mathbb{Z}\}\)

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.

Answer 1

The Correct Option is A

Step 1: Read the question as a determinant condition.
The two equations are homogeneous (right-hand sides are zero). Such a system fails to have a unique solution exactly when the coefficient determinant vanishes, because only then can a nontrivial solution exist alongside the trivial one.
Step 2: Write down the determinant.
The coefficient matrix is \[ A = \begin{bmatrix} \cos\theta & \sec\theta \\ \sin\theta & \tan\theta \end{bmatrix} \] so \[ \det A = \cos\theta\,\tan\theta - \sin\theta\,\sec\theta \]
Step 3: Clear everything over a common denominator $\cos\theta$.
Writing $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\sec\theta = \dfrac{1}{\cos\theta}$, \[ \det A = \sin\theta - \frac{\sin\theta}{\cos\theta} = \frac{\sin\theta\cos\theta - \sin\theta}{\cos\theta} \]
Step 4: Factor the numerator.
Pull out $\sin\theta$: \[ \det A = \frac{\sin\theta(\cos\theta - 1)}{\cos\theta} \] Here $\cos\theta \neq 0$ is guaranteed by the restriction $\theta \neq \dfrac{(2n+1)\pi}{2}$, so the denominator is safe.
Step 5: Set the determinant to zero.
A vanishing determinant requires the numerator to be zero: \[ \sin\theta(\cos\theta - 1) = 0 \] giving either $\sin\theta = 0$ or $\cos\theta = 1$.
Step 6: Merge the two cases.
$\sin\theta = 0$ gives $\theta = n\pi$. The case $\cos\theta = 1$ gives $\theta = 2n\pi$, which is already a subset of $\theta = n\pi$. So the complete solution set is $\{n\pi : n \in \mathbb{Z}\}$, exactly option (A).
\[ \boxed{\theta \in \{n\pi : n \in \mathbb{Z}\}} \]