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A value of $\theta$ in $\left(0, \frac{\pi}{2}\right)$ and satisfying $\left|\begin{array}{ccc}1+\sin ^2 \theta & \cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & 1+\cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta\end{array}\right|=0$
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Verified Answer
The correct answer is:
$\frac{7 \pi}{24}$
Given determinant
$\Delta=\left|\begin{array}{ccc}1+\sin ^2 \theta & \cos ^2 \theta & 4 \sin 4 \theta \\\sin ^2 \theta & 1+\cos ^2 \theta & 4 \sin 4 \theta \\\sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta
\end{array}\right|=0$
On applying $C_1 \rightarrow C_1+C_2+C_3$, we get
$\Delta=\left|\begin{array}{ccc}2+4 \sin 4 \theta & \cos ^2 \theta & 4 \sin 4 \theta \\2+4 \sin 4 \theta & 1+\cos ^2 \theta & 4 \sin 4 \theta \\2+4 \sin 4 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta
\end{array}\right|=0$
$\Rightarrow \Delta=(2+4 \sin 4 \theta)\left|\begin{array}{ccc}1 & \cos ^2 \theta & 4 \sin 4 \theta \\ 1 & 1+\cos ^2 \theta & 4 \sin 4 \theta \\ 1 & \cos ^2 \theta & 1+4 \sin 4 \theta\end{array}\right|=0$
On applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$, we get
$\Rightarrow \Delta=(1+2 \sin 4 \theta)\left|\begin{array}{ccc}
1 & \cos ^2 \theta & 4 \sin 4 \theta \\0 & 1 & 0 \\0 & 0 & 1
\end{array}\right|=0$
$\Rightarrow \Delta=2 \sin 4 \theta+1=0 \Rightarrow \sin 4 \theta=-\frac{1}{2} \Rightarrow 4 \theta=\frac{7 \pi}{6}$
$\Rightarrow \quad \theta=\frac{7 \pi}{24} \in\left(0, \frac{\pi}{2}\right)$
$\Delta=\left|\begin{array}{ccc}1+\sin ^2 \theta & \cos ^2 \theta & 4 \sin 4 \theta \\\sin ^2 \theta & 1+\cos ^2 \theta & 4 \sin 4 \theta \\\sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta
\end{array}\right|=0$
On applying $C_1 \rightarrow C_1+C_2+C_3$, we get
$\Delta=\left|\begin{array}{ccc}2+4 \sin 4 \theta & \cos ^2 \theta & 4 \sin 4 \theta \\2+4 \sin 4 \theta & 1+\cos ^2 \theta & 4 \sin 4 \theta \\2+4 \sin 4 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta
\end{array}\right|=0$
$\Rightarrow \Delta=(2+4 \sin 4 \theta)\left|\begin{array}{ccc}1 & \cos ^2 \theta & 4 \sin 4 \theta \\ 1 & 1+\cos ^2 \theta & 4 \sin 4 \theta \\ 1 & \cos ^2 \theta & 1+4 \sin 4 \theta\end{array}\right|=0$
On applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$, we get
$\Rightarrow \Delta=(1+2 \sin 4 \theta)\left|\begin{array}{ccc}
1 & \cos ^2 \theta & 4 \sin 4 \theta \\0 & 1 & 0 \\0 & 0 & 1
\end{array}\right|=0$
$\Rightarrow \Delta=2 \sin 4 \theta+1=0 \Rightarrow \sin 4 \theta=-\frac{1}{2} \Rightarrow 4 \theta=\frac{7 \pi}{6}$
$\Rightarrow \quad \theta=\frac{7 \pi}{24} \in\left(0, \frac{\pi}{2}\right)$
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