Question

The sum of all possible values of $\theta$ where $\theta \in (0,\frac{\pi }{2}),$ satisfying the equation $\left|\begin{array}{ccc}1+{\sin }^2\theta & {\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta & 1+{\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta & {\cos }^2\theta & 1+4\sin 4\theta \end{array}\right|=0,$ is

• A. $\frac{3\pi }{4}$
• B. $\frac{5\pi }{4}$
• C. $\frac{\pi }{4}$
• D. $\frac{11\pi }{12}$

Answer 1

The correct option is A $\frac{3\pi }{4}$

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

Applying $C_1\to C_1+C_2+C_3,$
$\Rightarrow \left|\begin{array}{ccc}2+4\sin 4\theta & {\cos }^2\theta & 4\sin 4\theta \\ 2+4\sin 4\theta & 1+{\cos }^2\theta & 4\sin 4\theta \\ 2+4\sin 4\theta & {\cos }^2\theta & 1+4\sin 4\theta \end{array}\right|=0$

Taking $2+4\sin 4\theta$ common from $C_1,$ then applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1,$
$\Rightarrow (2+4\sin 4\theta )\left|\begin{array}{ccc}1& {\cos }^2\theta & 4\sin 4\theta \\ 0& 1& 0\\ 0& 0& 1\end{array}\right|=0$
$\Rightarrow 2+4\sin 4\theta =0$
$\Rightarrow \sin 4\theta =-\frac{1}{2}$
As $\theta \in (0,\frac{\pi }{2}),$ we have $4\theta \in (0,2\pi )$
$\Rightarrow 4\theta =\frac{7\pi }{6},\frac{11\pi }{6}$
$\Rightarrow \theta =\frac{7\pi }{24},\frac{11\pi }{24}$