Statement-1: The system of linear equations $ \begin{aligned} & x+(\sin \alpha) y+(\cos \alpha) z=0 \\ &…
Statement-1: The system of linear equations
$
\begin{aligned}
& x+(\sin \alpha) y+(\cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& x-(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}
$
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 $\alpha$
$
\left|\begin{array}{ccc}
\cos \alpha & \sin \alpha & \cos \alpha \\
\sin \alpha & \cos \alpha & \sin \alpha \\
\cos \alpha & -\sin \alpha & -\cos \alpha
\end{array}\right|=0
$
has only one solution lying in the interval
$\left(0, \frac{\pi}{2}\right)$
Statement-1 is true, Statement-2 is true, Statement-2 is not correct explantion for Statement-1.
Statement-1 is true, Statement-2 is true, Statement-2 is a correct explantion for Statement-1.
Statement-1 is true, Statement- 2 is false.
Statememt-1 is false, Statement-2 is true.
Solution
$
\begin{aligned}
& \Rightarrow \alpha=\frac{64 x}{3 \times 16 x}=\frac{4}{3} \\
& \Delta_1=\left|\begin{array}{ccc}
1 & \sin \alpha & \cos \alpha \\
1 & \cos \alpha & \sin \alpha \\
1 & -\sin \alpha & \cos \alpha
\end{array}\right| \\
& =\left|\begin{array}{ccc}
0 & \sin \alpha-\cos \alpha & \cos \alpha-\sin \alpha \\
0 & \cos \alpha+\sin \alpha & \sin \alpha-\cos \alpha \\
1 & -\sin \alpha & \cos \alpha
\end{array}\right| \\
& =(\sin \alpha-\cos \alpha)^2-\left(\cos ^2 \alpha-\sin ^2 \alpha\right) \\
& =\sin ^2 \alpha+\cos ^2 \alpha-2 \sin \alpha \cdot \cos \alpha-\cos ^2 \alpha \\
& =2 \sin ^2 \alpha-2 \sin \alpha \cdot \cos \alpha \\
& =2 \sin ^2 \alpha(\sin \alpha-\cos \alpha)
\end{aligned}
$
Now, $\sin \alpha-\cos \alpha=0$ for only
$
\begin{aligned}
\alpha & =\frac{\pi}{4} \text { in }\left(0, \frac{\pi}{2}\right) \\
\therefore \quad \Delta_1 & =2(\sin \alpha) \times 0=0,
\end{aligned}
$
since value of $\sin \alpha$ is finite for $\alpha \in\left(0, \frac{\pi}{2}\right)$
Hence non-trivivial solution for only one value of $\alpha$ in $\left(0, \frac{\pi}{2}\right)$
$
\begin{aligned}
& \left|\begin{array}{ccc}
\cos \alpha & \sin \alpha & \cos \alpha \\
\sin \alpha & \cos \alpha & \sin \alpha \\
\cos \alpha & -\sin \alpha & -\cos \alpha
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
0 & \sin \alpha & \cos \alpha \\
0 & \cos \alpha & \sin \alpha \\
2 \cos \alpha & -\sin \alpha & -\cos \alpha
\end{array}\right|=0 \\
& \Rightarrow 2 \cos \alpha\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\
& \therefore \cos \alpha=0 \text { or } \sin ^2 \alpha-\cos ^2 \alpha=0
\end{aligned}
$
But $\cos \alpha=0$ not possible for any value of
$
\begin{aligned}
& \alpha \in\left(0, \frac{\pi}{2}\right) \\
& \therefore \sin ^2 \alpha-\cos ^2 \alpha=0 \Rightarrow \sin \alpha=-\cos \alpha,
\end{aligned}
$
which is also not possible for any value of
$
\alpha \in\left(0, \frac{\pi}{2}\right)
$
Hence, there is no solution.