Question

$\left| \begin{matrix} 1 + \sin^{2}\theta & \sin^{2}\theta & \sin^{2}\theta \\ \cos^{2}\theta & 1 + \cos^{2}\theta & \cos^{2}\theta \\ 4\sin 4\theta & 4\sin 4\theta & 1 + 4\sin 4\theta \end{matrix} \right| = 0$ then$\sin 4\theta$equal to.

• A. 1/2
• B. 1
• C. –1/2
• D. –1

Answer 1

Answer: (C)

–1/2

Answer 2

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

Using $C_{1} \rightarrow C_{1} - C_{2},C_{2} \rightarrow C_{2} - C_{3}$

⇒ $\left| \begin{matrix} 1 & 0 & \sin^{2}\theta \

• 1 & 1 & \cos^{2}\theta \ 0 & - 1 & 1 + 4\sin 4\theta \end{matrix} \right| = 0$

⇒ $2(1 + 2\sin 4\theta) = 0 \Rightarrow \sin 4\theta = \frac{- 1}{2}$.