Trigonometric Equations Question 89
Question: The value of $ \theta $ lying between 0 and $ \pi /2 $ and satisfying the equation $ | ,\begin{matrix} 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{matrix}, |=0 $
[IIT 1988; MNR 1992; Kurukshetra CEE 1998; DCE 1996]
Options:
A) $ \frac{7\pi }{24} $ or $ \frac{11\pi }{24} $
B) $ \frac{5\pi }{24} $
C) $ \frac{\pi }{24} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- The given determinant (Applying $ R_1\to R_1-R_3 $ and $ R_2\to R_2-R_3 $ ) reduces to $ | ,\begin{matrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ {{\sin }^{2}}\theta & {{\cos }^{2}}\theta & 1+4\sin 4\theta \\ \end{matrix}, |,=0 $
$ \Rightarrow $ $ 1+4\sin 4\theta +{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =0 $ (By expanding along $ R_1) $
Þ $ 4\sin 4\theta =-2 $
Þ $ \sin 4\theta =\frac{-1}{2} $
Þ $ 4\theta =\frac{7\pi }{6} $ or $ \frac{11\pi }{6} $ , ( $ 0<4\theta <2\pi $ ) Since, $ 0<\theta <\frac{\pi }{2} $
Þ $ 0<4\theta <2\pi $
Þ $ \theta =\frac{7\pi }{24},\frac{11\pi }{24} $
BETA
