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

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

Answer 1

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

$\left|\begin{array}{ccc}1+{\sin }^{2}A& {\cos }^{2}A& 2\sin 4\theta \\ {\sin }^{2}A& 1+{\cos }^{2}A& 2\sin 4\theta \\ {\sin }^{2}A& {\cos }^{2}A& 1+2\sin 4\theta \end{array}\right|=0$
$C_1\to C_1+C_2$
$\Rightarrow \left|\begin{array}{ccc}2& {\cos }^{2}A& 2\sin 4\theta \\ 2& 1+{\cos }^{2}A& 2\sin 4\theta \\ 1& {\cos }^{2}A& 1+2\sin 4\theta \end{array}\right|=0$
$R_1\to R_1-R_2$
$\Rightarrow \left|\begin{array}{ccc}0& -1& 0\\ 2& 1+{\cos }^{2}A& 2\sin 4\theta \\ 1& {\cos }^{2}A& 1+2\sin 4\theta \end{array}\right|=0$
$\Rightarrow 2(1+2\sin 4\theta )-2\sin 4\theta =0$
$\Rightarrow 1+\sin 4\theta =0$
$\sin 4\theta =-1$
$4\theta =-\frac{\pi }{2}$ or $4\theta =\frac{3\pi }{2}$
$\theta =-\frac{\pi }{8}$ or $\theta =\frac{3\pi }{8}\&A\in R$