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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Determinants
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The value of $\theta$ in the first quadrant satisfying the equation $\begin{vmatrix}1+\cos^2\theta&\sin^2\theta&4\sin 4\theta\\\cos^2\theta&1+\sin^2\theta&4\sin4\theta\\\cos^2\theta&\sin^2\theta&1+4\sin 4\theta\end{vmatrix}=0$ is

$\begin{array}{1 1}(A)\;\large\frac{5\pi}{24}&(B)\;\large\frac{7\pi}{24},\frac{5\pi}{24}\\(C)\;\large\frac{\pi}{24}&(D)\;\large\frac{11\pi}{6},\frac{7\pi}{6}\end{array}$

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1 Answer

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Operating $R_2-R_1$ and $R_3-R_2$ we get
$\begin{vmatrix}1+\cos^2\theta&\sin ^2\theta&4\sin 4\theta\\-1&1&0\\0&-1&1\end{vmatrix}=0$
Operating $C_1=C_1+C_2+C_3$
$\begin{vmatrix}2+4\sin 4\theta&\sin ^2\theta&4\sin 4\theta\\0&1&0\\0&-1&1\end{vmatrix}=0$
$\Rightarrow 2+4\sin 4\theta=0$
$\Rightarrow \sin 4\theta=-\large\frac{1}{2}$
$\Rightarrow -\sin \large\frac{\pi}{6}$
$\Rightarrow \sin \large\frac{7\pi}{6}$$,\sin \large\frac{11\pi}{6}$
Hence (D) is the correct answer.
answered Apr 23, 2014 by sreemathi.v
 

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