Question

Assertion :The system of linear equations
$x+\left(\sin \theta \right)y+\left(\cos \theta \right)z=0$
$x+\left(\cos \theta \right)y+\left(\sin \theta \right)z=0$
$x-\left(\sin \theta \right)y-\left(\cos \theta \right)z=0$
has a non-trivial solution for only one value of $\theta$ lying in the interval $(0,\frac{\pi }{2})$. Reason: The equation in $\theta$ $\left|\begin{array}{ccc}\cos \theta & \sin \theta & \cos \theta \\ \sin \theta & \cos \theta & \sin \theta \\ \cos \theta & -\sin \theta & -\cos \theta \end{array}\right|=0$ has only one solution lying in the interval $(0,\frac{\pi }{2})$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Statement-$1$: For non-trivial solution, put $\Delta =0$
$\Rightarrow \left|\begin{array}{ccc}1& \sin \theta & \cos \theta \\ 1& \cos \theta & \sin \theta \\ 1& \sin \theta & -\cos \theta \end{array}\right|=0$
$\Rightarrow 1(-{\cos }^2\theta +{\sin }^2\theta )-\sin \theta (-\cos \theta -\sin \theta )+\cos \theta (\sin \theta -\cos \theta )=\theta$
$\Rightarrow (\sin \theta +\cos \theta )(\sin \theta -\cos \theta +\sin \theta -\cos \theta )$
$\Rightarrow 2(\sin \theta +\cos \theta )\cdot (\sin \theta -\cos \theta )=0$
$\therefore \tan \theta =-1,1\Rightarrow \theta =\frac{\pi }{4}\in (0,\frac{\pi }{2})$.
Statement-$2$: We have,
$\left|\begin{array}{ccc}\cos \theta & \sin \theta & \cos \theta \\ \sin \theta & \cos \theta & \sin \theta \\ \cos \theta & -\sin \theta & -\cos \theta \end{array}\right|=0$
$\Rightarrow \cos \theta (-{\cos }^2\theta +{\sin }^2\theta )-\sin \theta (-2\sin \theta \cdot \cos \theta )+\cos \theta (-1)=0$
$\Rightarrow -\cos \theta \cdot \cos 2\theta +\sin \theta \cdot \sin 2\theta =\cos \theta$
$\Rightarrow \cos \theta +\cos 3\theta =0$
$\Rightarrow 2\cos 2\theta \cdot \cos \theta =0$
$\therefore \cos \theta =0$ or $\cos 2\theta =0$
$\Rightarrow \theta =\frac{\pi }{4}\in (0,\frac{\pi }{2})$.