Question:
Statement-1: The system of linear equations
\(x+(\sin\alpha)y+(\cos\alpha)z=0\)
\(x+(\cos\alpha)y+(\sin\alpha)z=0\)
\(x-(\sin\alpha)y-(\cos\alpha)z=0\)
has a non-trivial solution for only one value of \(\alpha\) lying in the interval \(\left(0, \frac{\pi}{2}\right)\).
Statement-2: The equation in \(\alpha\)
\(\begin{vmatrix}\cos\alpha & \sin\alpha & \cos\alpha\\ \sin\alpha & \cos\alpha & \sin\alpha\\ \cos\alpha & -\sin\alpha & -\cos\alpha\end{vmatrix}=0\)
has only one solution lying in the interval \(\left(0, \frac{\pi}{2}\right)\)
Statement-1: The system of linear equations
\(x+(\sin\alpha)y+(\cos\alpha)z=0\)
\(x+(\cos\alpha)y+(\sin\alpha)z=0\)
\(x-(\sin\alpha)y-(\cos\alpha)z=0\)
has a non-trivial solution for only one value of \(\alpha\) lying in the interval \(\left(0, \frac{\pi}{2}\right)\).
Statement-2: The equation in \(\alpha\)
\(\begin{vmatrix}\cos\alpha & \sin\alpha & \cos\alpha\\ \sin\alpha & \cos\alpha & \sin\alpha\\ \cos\alpha & -\sin\alpha & -\cos\alpha\end{vmatrix}=0\)
has only one solution lying in the interval \(\left(0, \frac{\pi}{2}\right)\)
\(x+(\sin\alpha)y+(\cos\alpha)z=0\)
\(x+(\cos\alpha)y+(\sin\alpha)z=0\)
\(x-(\sin\alpha)y-(\cos\alpha)z=0\)
has a non-trivial solution for only one value of \(\alpha\) lying in the interval \(\left(0, \frac{\pi}{2}\right)\).
Statement-2: The equation in \(\alpha\)
\(\begin{vmatrix}\cos\alpha & \sin\alpha & \cos\alpha\\ \sin\alpha & \cos\alpha & \sin\alpha\\ \cos\alpha & -\sin\alpha & -\cos\alpha\end{vmatrix}=0\)
has only one solution lying in the interval \(\left(0, \frac{\pi}{2}\right)\)
Show Hint
A homogeneous system has a non-trivial solution iff its coefficient determinant is 0; evaluate both determinants.
Updated On: Oct 31, 2026
- Statement-1 is true, Statement-2 is true, Statement-2 is not correct explanation for Statement-1.
- Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.
- Statement-1 is true, Statement-2 is false.
- Statement-1 is false, Statement-2 is true.
Show Solution
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The Correct Option is A
Solution and Explanation
Statement-1: a non-trivial solution exists iff the coefficient determinant vanishes. \(\begin{vmatrix}1 & \sin\alpha & \cos\alpha\\ 1 & \cos\alpha & \sin\alpha\\ 1 & -\sin\alpha & -\cos\alpha\end{vmatrix}\) \(=(\sin^2\alpha-\cos^2\alpha)-\sin\alpha(-\cos\alpha-\sin\alpha)+\cos\alpha(-\sin\alpha-\cos\alpha)=-\cos2\alpha+\sin^2\alpha-\cos^2\alpha=-2\cos2\alpha\).
\(-2\cos2\alpha=0 \Rightarrow \alpha=\pi/4\) in \(\left(0,\frac{\pi}{2}\right)\). Exactly one value, so Statement-1 is true.
Statement-2: expanding the given determinant gives \(\cos\alpha(\sin^2\alpha-\cos^2\alpha)-\sin\alpha(-2\sin\alpha\cos\alpha)+\cos\alpha(-1)=-2\cos\alpha\cos2\alpha\).
In \(\left(0,\frac{\pi}{2}\right)\), \(\cos\alpha\neq0\), so the only root is \(\cos2\alpha=0\), i.e. \(\alpha=\pi/4\). Statement-2 is also true.
The determinant in Statement-2 is not the coefficient determinant of the given system (its first column is cos, sin, cos instead of 1, 1, 1), so it is not the condition that explains Statement-1; it merely gives the same root. Hence Statement-2 is not a correct explanation.
Answer: (1) Statement-1 is true, Statement-2 is true, Statement-2 is not correct explanation for Statement-1.
\(-2\cos2\alpha=0 \Rightarrow \alpha=\pi/4\) in \(\left(0,\frac{\pi}{2}\right)\). Exactly one value, so Statement-1 is true.
Statement-2: expanding the given determinant gives \(\cos\alpha(\sin^2\alpha-\cos^2\alpha)-\sin\alpha(-2\sin\alpha\cos\alpha)+\cos\alpha(-1)=-2\cos\alpha\cos2\alpha\).
In \(\left(0,\frac{\pi}{2}\right)\), \(\cos\alpha\neq0\), so the only root is \(\cos2\alpha=0\), i.e. \(\alpha=\pi/4\). Statement-2 is also true.
The determinant in Statement-2 is not the coefficient determinant of the given system (its first column is cos, sin, cos instead of 1, 1, 1), so it is not the condition that explains Statement-1; it merely gives the same root. Hence Statement-2 is not a correct explanation.
Answer: (1) Statement-1 is true, Statement-2 is true, Statement-2 is not correct explanation for Statement-1.
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