Question

If $\left|\begin{array}{ccc}1+{\sin }^2& {\cos }^2\theta & 4\sin 2\theta \\ {\sin }^2& 1+{\cos }^2& 4\sin 2\theta \\ {\sin }^2\theta & {\cos }^2\theta & 4\sin 2\theta -1\end{array}\right|=0$ and $0<\theta <\frac{\pi }{2}$ then $\cos 4\theta$

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $0$
• D. $\frac{-1}{2}$

Answer 1

Answer: (A)

Given, $\left|\begin{array}{ccc}1+{\sin }^2\theta & {\cos }^2\theta & 4\sin 2\theta \\ {\sin }^2\theta & 1+{\cos }^2\theta & 4\sin 2\theta \\ {\sin }^2\theta & {\cos }^2\theta & 4\sin 2\theta -1\end{array}\right|=0$
Applying $C_1\to C_1+C_2$
$\Rightarrow \left|\begin{array}{ccc}2& {\cos }^2\theta & 4\sin 2\theta \\ 2& 1+{\cos }^2\theta & 4\sin 2\theta \\ 1& {\cos }^2\theta & 4\sin 2\theta -1\end{array}\right|=0$
Applying $R_2\to R_2-R_1,R_3\to 2R_3-R_1$
$\Rightarrow \left|\begin{array}{ccc}2& {\cos }^2\theta & 4\sin 2\theta \\ 0& 1& 0\\ 0& {\cos }^2\theta & 4\sin 2\theta -2\end{array}\right|=0$
$\Rightarrow 2(4\sin 2\theta -2-0)=0$
$\Rightarrow \sin 2\theta =\frac{1}{2}$
Now, $\cos 4\theta =1-2{\sin }^{2}2\theta$
$=1-2{\left(\frac{1}{2}\right)}^2=\frac{1}{2}$