Question

The value of $\theta$ lying between $-\frac{\pi}{4} \& \frac{\pi}{2}$ and $0 \leq A \leq \frac{\pi}{2}$ and satisfying the equation $\begin{vmatrix}1+\sin ^2 A & \cos ^2 A & 2 \sin 4 \theta \\ \sin ^2 A & 1+\cos ^2 A & 2 \sin 4 \theta \\ \sin ^2 A & \cos ^2 A & 1+2 \sin 4 \theta\end{vmatrix}=0$ 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)

$\begin{vmatrix}1+\sin ^2 A & \cos ^2 A & 2 \sin 4 \theta \\ \sin ^2 A & 1+\cos ^2 A & 2 \sin 4 \theta \\ \sin ^2 A & \cos ^2 A & 1+2 \sin 4 \theta\end{vmatrix}=0$
$C_1 \rightarrow C_1+C_2$
$\Rightarrow\begin{vmatrix}2 & \cos ^2 A & 2 \sin 4 \theta \\ 2 & 1+\cos ^2 A & 2 \sin 4 \theta \\ 1 & \cos ^2 A & 1+2 \sin 4 \theta\end{vmatrix}=0$
$R _1 \rightarrow R _1- R _2$
$\Rightarrow\begin{vmatrix}0 & -1 & 0 \\ 2 & 1+\cos ^2 A & 2 \sin 4 \theta \\ 1 & \cos ^2 A & 1+2 \sin 4 \theta\end{vmatrix}=0$
$\Rightarrow 2(1+2 \sin 4 \theta)-2 \sin 4 \theta=0$
$\Rightarrow 1+\sin 4 \theta=0$
$\sin 4 \theta=-1$
$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$