Trigonometrical Equations 1 Question 20
20. The values of $\theta$ lying between $\theta=0$ and $\theta=\pi / 2$ and satisfying the equation
$ \left|\begin{array}{ccc} 1+\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 4 \theta \\ \sin ^{2} \theta & 1+\cos ^{2} \theta & 4 \sin 4 \theta \\ \sin ^{2} \theta & \cos ^{2} \theta & 1+4 \sin 4 \theta \end{array}\right|=0, \text { is } $
(a) $7 \pi / 24$
(b) $5 \pi / 24$
(c) $11 \pi / 24$
(d) $\pi / 24$
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Answer:
Correct Answer: 20. (a, c)
Solution:
- Given, $\left|\begin{array}{ccc}1+\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 4 \theta \\ \sin ^{2} \theta & 1+\cos ^{2} \theta & 4 \sin 4 \theta \\ \sin ^{2} \theta & \cos ^{2} \theta & 1+4 \sin 4 \theta\end{array}\right|=0$
Applying $R _3 \rightarrow R _3-R _1$ and $R _2 \rightarrow R _2-R _1$, we get
$ \left|\begin{array}{ccc} 1+\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 4 \theta \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right|=0 $
Applying $C _1 \rightarrow C _1+C _2$, we get
$ \begin{array}{cc} & \left|\begin{array}{ccc} 2 & \cos ^{2} \theta & 4 \sin 4 \theta \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right|=0 \\ \Rightarrow & 2+4 \sin 4 \theta=0 \\ \Rightarrow & \sin 4 \theta=\frac{-1}{2} \\ \Rightarrow & 4 \theta=n \pi+(-1)^{n}-\frac{\pi}{6} \\ \Rightarrow & \theta=\frac{n \pi}{4}+(-1)^{n+1} \frac{\pi}{24} \end{array} $
Clearly, $\theta=\frac{7 \pi}{24}, \frac{11 \pi}{24}$ are two values of $\theta$ lying between 0 and $\pi / 2$.
