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Mathematics Question on Determinants

If ∣1+sin⁡2cos⁡2θ4sin⁡2θ sin⁡21+cos⁡24sin⁡2θ sin⁡2θcos⁡2θ4sin⁡2θ−1 ∣=0\begin{vmatrix} {1+\sin^2}&{\cos^2 \theta } &{4\sin 2\theta}\\\ {\sin^2}&{1+\cos^2}& {4\sin2\theta} \\\ {\sin^2\theta}&{\cos^2\theta}&{4\sin2\theta-1}\\\ \end{vmatrix} =0 and $ 0< \theta

A

12\frac {1}{2}

B

32\frac {\sqrt {3}}{2}

C

00

D

−12\frac {-1}{2}

Answer

12\frac {1}{2}

Explanation

Solution

Given, ∣1+sin⁡2θcos⁡2θ4sin⁡2θ sin⁡2θ1+cos⁡2θ4sin⁡2θ sin⁡2θcos⁡2θ4sin⁡2θ−1∣=0\begin{vmatrix}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{vmatrix}=0
Applying C1→C1+C2C_{1} \rightarrow C_{1}+C_{2}
⇒∣2cos⁡2θ4sin⁡2θ 21+cos⁡2θ4sin⁡2θ 1cos⁡2θ4sin⁡2θ−1∣=0\Rightarrow\begin{vmatrix}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{vmatrix}=0
Applying R2→R2−R1,R3→2R3−R1R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow 2 R_{3}-R_{1}
⇒∣2cos⁡2θ4sin⁡2θ 010 0cos⁡2θ4sin⁡2θ−2∣=0\Rightarrow\begin{vmatrix}2 & \cos ^{2} \theta & 4 \sin 2 \theta \\\ 0 & 1 & 0 \\\ 0 & \cos ^{2} \theta & 4 \sin 2 \theta-2\end{vmatrix}=0
⇒     2(4sin⁡2θ−2−0)=0\Rightarrow\,\,\,\,\,2(4 \sin 2 \theta-2-0)=0
⇒     sin⁡2θ=12\Rightarrow\,\,\,\,\,\sin 2 \theta=\frac{1}{2}
Now, cos⁡4θ=1−2sin⁡22θ\cos 4 \theta=1-2 \sin ^{2} 2 \theta
=1−2(12)2=12=1-2\left(\frac{1}{2}\right)^{2}=\frac{1}{2}