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The value of $\theta $ lying between $0$ and $\pi /2$ and satisfying the equation
$\left| {\,\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\cos }^2}\theta }&{4\sin 4\theta }\\{{{\sin }^2}\theta }&{1 + {{\cos }^2}\theta }&{4\sin 4\theta }\\{{{\sin }^2}\theta }&{{{\cos }^2}\theta }&{1 + 4\sin 4\theta }\end{array}\,} \right| = 0$
$\frac{{7\pi }}{{24}}$ or $\frac{{11\pi }}{{24}}$
$\frac{{5\pi }}{{24}}$
$\frac{\pi }{{24}}$
None of these
Solution
(a) The given determinant
(Applying ${R_1} \to {R_1} – {R_3}$ and ${R_2} \to {R_2} – {R_3}$) reduces to
$\left| {\,\begin{array}{*{20}{c}}1&0&{ – 1}\\0&1&{ – 1}\\{{{\sin }^2}\theta }&{{{\cos }^2}\theta }&{1 + 4\sin 4\theta }\end{array}\,} \right|\, = 0$
$ \Rightarrow $ $1 + 4\sin 4\theta + {\cos ^2}\theta + {\sin ^2}\theta = 0$
(By expanding along ${R_1})$
==> $4\sin 4\theta = – 2$ ==> $\sin 4\theta = \frac{{ – 1}}{2}$
==> $4\theta = \frac{{7\pi }}{6}$ or $\frac{{11\pi }}{6}$, ($0 < 4\theta < 2\pi $)
Since, $0 < \theta < \frac{\pi }{2}$
==> $0 < 4\theta < 2\pi $
==> $\theta = \frac{{7\pi }}{{24}},\,\,\frac{{11\pi }}{{24}}$.
