Question

The values of $\theta$ lying between $\theta=0$ and $\theta=\pi / 2$ and satisfying the equation $\Delta=\begin{vmatrix} 1+\cos ^2 \theta & \sin ^2 \theta & 4 \sin 4 \theta \\ \cos ^2 \theta & 1+\sin ^2 \theta & 4 \sin 4 \theta \\ \cos ^2 \theta & \sin ^2 \theta & 1+4 \sin 4 \theta \end{vmatrix}=0$ are given by

• A. $\pi / 24,5 \pi / 24$
• B. $7 \pi / 24,11 \pi / 24$
• C. $5 \pi / 24,7 \pi / 24$
• D. $11 \pi / 24, \pi / 24$

Answer 1

Answer: (B)

$C_3 \rightarrow C_3+C_1+C_2$, gives $\Delta=(2+4 \sin 4 \theta)\begin{vmatrix} 1+\cos ^2 \theta & \sin ^2 \theta & 1 \\ \cos ^2 \theta & 1+\sin ^2 \theta & 1 \\ \cos ^2 \theta & \sin ^2 \theta & 1 \end{vmatrix}$
Applying $R_3 \rightarrow R_3-R_2, R_2 \rightarrow R_2-R_1$, we get $\Delta =2(1+2 \sin 4 \theta)\begin{vmatrix} 1+\cos ^2 \theta & \sin ^2 \theta & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 0 \end{vmatrix}$
$=2(1+2 \sin 4 \theta)$
$\Delta =0 \Rightarrow \sin 4 \theta=-1 / 2$
Now, $0 \leq \theta \leq \frac{\pi}{2} \Rightarrow 0 \leq 4 \theta \leq 2 \pi$
$\therefore 4 \theta=\frac{7 \pi}{6}, \frac{11 \pi}{6} \Rightarrow \theta=\frac{7 \pi}{24}, \frac{11 \pi}{24}$