Question

$\left|\begin{array}{ccc}1+{\sin }^2\theta & {\sin }^2\theta & si{n}^2\theta \\ co{s}^2\theta & 1+co{s}^2\theta & co{s}^2\theta \\ 4\sin 4\theta & 4\sin 4\theta & 1+4\sin 4\theta \end{array}\right|=0$, then $\sin 4\theta$ equals to.

• A. $1/2$
• B. $1$
• C. $-1/2$
• D. $-1$

Answer 1

The correct option is C $-1/2$

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

$C_1\to C_1-C_2$

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

$R_1\to R_1-R_2$

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

$0-(-1)(2+8\sin 4\theta -4\sin 4\theta )+0=0$

$2+8\sin 4\theta -4\sin 4\theta =0$

$2+4\sin 4\theta (2-1)=0$

$4\sin 4\theta (2-1)=-2$

$\sin 4\theta (2-1)=-\frac{1}{2}$

$\sin 4\theta =-\frac{1}{2}$

Option C