If $\Delta_{1}=\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ and
$\Delta_{2}=\left|\begin{array}{ccc}\mathrm{x} & \sin 2 \theta & \cos 2 \theta \\ -\sin 2 \theta & -\mathrm{x} & 1 \\ \cos 2 \theta & 1 & \mathrm{x}\end{array}\right|, \mathrm{x} \neq 0 ;$ then for
all $\theta \in\left(0, \frac{\pi}{2}\right):$
Correct Option: , 2
$\Delta_{1}=f(\theta)=\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|=-x^{3}$c
and $\Delta_{2}=f(2 \theta)=\left|\begin{array}{ccc}x & \sin 2 \theta & \cos 2 \theta \\ -\sin 2 \theta & -x & 1 \\ \cos 2 \theta & 1 & x\end{array}\right|=-x^{3}$
So $\Delta_{1}+\Delta_{2}=-2 x^{3}$
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