A) \[\frac{\sqrt{3}}{2}\]
B) \[0\]
C) \[\frac{-1}{2}\]
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
Given, \[\left| \begin{matrix} 1+{{\sin }^{2}}\theta & {{\cos }^{2}}\theta & 4\sin 2\theta \\ {{\sin }^{2}}\theta & 1+{{\cos }^{2}}\theta & 4\sin 2\theta \\ {{\sin }^{2}}\theta & {{\cos }^{2}}\theta & 4\sin 2\theta -1 \\ \end{matrix} \right|=0\] Applying \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}\] \[\Rightarrow \] \[\left| \begin{matrix} 2 & {{\cos }^{2}}\theta & 4\sin 2\theta \\ 2 & 1+{{\cos }^{2}}\theta & 4\sin 2\theta \\ 1 & {{\cos }^{2}}\theta & 4\sin \,2\theta -1 \\ \end{matrix} \right|=0\] Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},{{R}_{3}}\to 2{{R}_{3}}-{{R}_{1}}\] \[\Rightarrow \] \[\left| \begin{matrix} 2 & {{\cos }^{2}}\theta & 4\sin 2\theta \\ 0 & 1 & 0 \\ 0 & {{\cos }^{2}}\theta & 4\sin 2\theta -2 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[2(4\sin 2\theta -2-0)=0\] \[\Rightarrow \] \[\sin 2\theta =\frac{1}{2}\] Now, \[\cos 4\theta =1-2{{\sin }^{2}}2\theta \] \[=1-2{{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{2}\]
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