# If the solve the problem

Question:

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 $\theta \in\left(0, \frac{\pi}{2}\right):$

1. $\Delta_{1}-\Delta_{2}=\mathrm{x}(\cos 2 \theta-\cos 4 \theta)$

2. $\Delta_{1}+\Delta_{2}=-2 \mathrm{x}^{3}$

3. $\Delta_{1}-\Delta_{2}=-2 \mathrm{x}^{3}$

4. $\Delta_{1}+\Delta_{2}=-2\left(\mathrm{x}^{3}+\mathrm{x}-1\right)$

Correct Option: , 2

Solution:

$\Delta_{1}=f(\theta)=\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|=-x^{3}$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 $\Delta_{1}+\Delta_{2}=-2 x^{3}$