If \(\left|\begin{array}{ccc}1+\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & 1+\cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta-1\end{array}\right|=0\) and \(0 < \theta < \frac{\pi}{2}\), then \(\cos 4 \theta\) is equal to

Given,
\[
\left|\begin{array}{ccc}
1+\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta \\
\sin ^{2} \theta & 1+\cos ^{2} \theta & 4 \sin 2 \theta \\
\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta-1
\end{array}\right|=0
\]
Applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}\)
\[
\Rightarrow\left|\begin{array}{ccc}
2 & \cos ^{2} \theta & 4 \sin 2 \theta \\
2 & 1+\cos ^{2} \theta & 4 \sin 2 \theta \\
1 & \cos ^{2} \theta & 4 \sin 2 \theta-1
\end{array}\right|=0
\]
Applying \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow 2 \mathrm{R}_{3}-\mathrm{R}_{1}\)
\[
\begin{aligned}
&\Rightarrow \quad\left|\begin{array}{ccc}
2 & \cos ^{2} \theta & 4 \sin 2 \theta \\
0 & 1 & 0 \\
0 & \cos ^{2} \theta & 4 \sin 2 \theta-2
\end{array}\right|=0 \\
&\Rightarrow \quad 2(4 \sin 2 \theta-2-0)=0 \\
&\Rightarrow \quad \sin 2 \theta=\frac{1}{2} \\
&\text { Now, } \cos 4 \theta=1-2 \sin ^{2} 2 \theta \\
&=1-2\left(\frac{1}{2}\right)^{2}=\frac{1}{2}
\end{aligned}
\]