Question

If $\begin{vmatrix} {1+\sin^2}&{\cos^2 \theta } &{4\sin 2\theta}\\ {\sin^2}&{1+\cos^2}& {4\sin2\theta} \\ {\sin^2\theta}&{\cos^2\theta}&{4\sin2\theta-1}\\ \end{vmatrix} =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, $\begin{vmatrix}1+\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & 1+\cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta-1\end{vmatrix}=0$
Applying $C_{1} \rightarrow C_{1}+C_{2}$
$\Rightarrow \begin{vmatrix}2 & \cos ^{2} \theta & 4 \sin 2 \theta \\ 2 & 1+\cos ^{2} \theta & 4 \sin 2 \theta \\ 1 & \cos ^{2} \theta & 4 \sin 2 \theta-1\end{vmatrix}=0$
Applying $R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow 2 R_{3}-R_{1}$
$\Rightarrow \begin{vmatrix}2 & \cos ^{2} \theta & 4 \sin 2 \theta \\ 0 & 1 & 0 \\ 0 & \cos ^{2} \theta & 4 \sin 2 \theta-2\end{vmatrix}=0$
$\Rightarrow \,\,\,\,\,2(4 \sin 2 \theta-2-0)=0$
$\Rightarrow \,\,\,\,\,\sin 2 \theta=\frac{1}{2}$
Now, $\cos 4 \theta=1-2 \sin ^{2} 2 \theta$
$=1-2\left(\frac{1}{2}\right)^{2}=\frac{1}{2}$