If left|begin{array}{ccc}1+sin ^{2} theta & c | vedclass

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(D) Given the determinant equation:$\left|\begin{array}{ccc}1+\sin ^{2} 	heta & \cos ^{2} 	heta & 4 \sin 2 	heta \ \sin ^{2} 	heta & 1+\cos ^{2}

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$\frac{1}{2}$

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(D) Given the determinant equation:$\left|\begin{array}{ccc}1+\sin ^{2} 	heta & \cos ^{2} 	heta & 4 \sin 2 	heta \ \sin ^{2} 	heta & 1+\cos ^{2} 	heta & 4 \sin 2 	heta \ \sin ^{2} 	heta & \cos ^{2} 	heta & 4 \sin 2 	heta-1\end{array}ight|=0$Applying the column operation $C_{1} ightarrow C_{1}+C_{2}$:$\left|\begin{array}{ccc}1+\sin ^{2} 	heta+\cos ^{2} 	heta & \cos ^{2} 	heta & 4 \sin 2 	heta \ \sin ^{2} 	heta+1+\cos ^{2} 	heta & 1+\cos ^{2} 	heta & 4 \sin 2 	heta \ \sin ^{2} 	heta+\cos ^{2} 	heta & \cos ^{2} 	heta & 4 \sin 2 	heta-1\end{array}ight|=0$Since $\sin ^{2} 	heta+\cos ^{2} 	heta=1$,this simplifies to:$\left|\begin{array}{ccc}2 & \cos ^{2} 	heta & 4 \sin 2 	heta \ 2 & 1+\cos ^{2} 	heta & 4 \sin 2 	heta \ 1 & \cos ^{2} 	heta & 4 \sin 2 	heta-1\end{array}ight|=0$Applying row operations $R_{2} ightarrow R_{2}-R_{1}$ and $R_{3} ightarrow 2 R_{3}-R_{1}$:$\left|\begin{array}{ccc}2 & \cos ^{2} 	heta & 4 \sin 2 	heta \ 0 & 1 & 0 \ 0 & \cos ^{2} 	heta & 4 \sin 2 	heta-2\end{array}ight|=0$Expanding along the first column:$2 	imes [1 	imes (4 \sin 2 	heta-2) - 0] = 0$$8 \sin 2 	heta-4=0 \Rightarrow \sin 2 	heta=\frac{1}{2}$Now,using the identity $\cos 4 	heta=1-2 \sin ^{2} 2 	heta$:$\cos 4 	heta=1-2\left(\frac{1}{2}ight)^{2}=1-2\left(\frac{1}{4}ight)=1-\frac{1}{2}=\frac{1}{2}$

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

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