A) 1/2
B) 1
C) -1/2
D) -1
Correct Answer: C
Solution :
\[\left| \,\begin{matrix} 1+{{\sin }^{2}}\theta & {{\sin }^{2}}\theta & {{\sin }^{2}}\theta \\ {{\cos }^{2}}\theta & 1+{{\cos }^{2}}\theta & {{\cos }^{2}}\theta \\ 4\sin 4\theta & 4\sin 4\theta & 1+4\sin 4\theta \\ \end{matrix}\, \right|=0\]Using \[{{C}_{1}}\to {{C}_{1}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}\] Þ \[\left| \,\begin{matrix} 1 & 0 & {{\sin }^{2}}\theta \\ -1 & 1 & {{\cos }^{2}}\theta \\ 0 & -1 & 1+4\sin 4\theta \\ \end{matrix}\, \right|=0\] Þ \[2\,(1+2\sin 4\theta )=0\Rightarrow \sin 4\theta =\frac{-1}{2}\].You need to login to perform this action.
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