Question

The value of $\theta$ lying between $-\frac{\pi }{4}$ & $\frac{\pi }{2}$ and $0\le A\le \frac{\pi }{2}$ and satisfying the equation $\left|\begin{array}{ccc}1+si{n}^{2}A& co{s}^{2}A& 2sin4\theta \\ si{n}^{2}A& 1+co{s}^{2}A& 2sin4\theta \\ si{n}^{2}A& co{s}^{2}A& 1+2sin4\theta \end{array}\right|=0$ are

• A. $A=\frac{\pi }{4},\theta =-\frac{\pi }{8}$
• B. $A=\frac{\pi }{4},\theta =0$
• C. $A=\frac{\pi }{5},\theta =-\frac{\pi }{8}$
• D. $A=\frac{\pi }{6},\theta =-\frac{3\pi }{8}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $A=\frac{\pi }{4},\theta =-\frac{\pi }{8}$
C $A=\frac{\pi }{5},\theta =-\frac{\pi }{8}$
D $A=\frac{\pi }{6},\theta =-\frac{3\pi }{8}$

$\left|\begin{array}{ccc}1+{\sin }^{2}A& {\cos }^{2}A& 2\sin 4\theta \\ {\sin }^{2}A& 1+{\cos }^{2}A& 2\sin 4\theta \\ {\sin }^{2}A& {\cos }^{2}A& 1+2\sin 4\theta \end{array}\right|=0$

Applying $C_1\to C_1+C_2$

$\left|\begin{array}{ccc}2& {\cos }^{2}A& 2\sin 4\theta \\ 2& 1+{\cos }^{2}x& 2\sin 4\theta \\ 1& {\cos }^{2}A& 1+2\sin 4\theta \end{array}\right|=0$

Applying $R_2\to R_2-R_1,$ we get

$\left|\begin{array}{ccc}2& {\cos }^{2}A& 2\sin 4\theta \\ 0& 1& 0\\ 1& {\cos }^{2}A& 1+2\sin 4\theta \end{array}\right|=0$

Expanding along $R_2,$ we get

$2(1+2\sin 4\theta )-2\sin 4\theta =0\Rightarrow 2+4\sin 4\theta -2\sin 4\theta =0$

$\Rightarrow 2+2\sin 4\theta =0\Rightarrow \sin 4\theta =-1\Rightarrow 4\theta =n\pi +{(-1)}^n\left(\frac{3\pi }{2}\right)$

$\Rightarrow \theta =\frac{n\pi }{4}+{(-1)}^n\left(\frac{3\pi }{8}\right)$

Hence $A$ can be anything and $\theta =-\frac{\pi }{8},-3\frac{\pi }{8},...$