(A) ∆=2(1−cos²x)
(B) ∆=2(2−sin²x)
(C)Minimum value of ∆ is 2
(D)Maximum value of ∆ is 4

Choosethecorrectanswerfromtheoptionsgivenbelow:

The correct answer is (B), (C), and (D) only.

Step-by-Step Explanation:

From the given image, the determinant $\Delta$ is represented as:
$$\Delta =\left|\begin{array}{ccc}1& \cos x& 1\\ -\cos x& 1& \cos x\\ -1& -\cos x& 1\end{array}\right|$$

We evaluate the determinant by expanding along the first row:
$$\Delta =1·\left|\begin{array}{cc}1& \cos x\\ -\cos x& 1\end{array}\right|-\cos x·\left|\begin{array}{cc}-\cos x& \cos x\\ -1& 1\end{array}\right|+1·\left|\begin{array}{cc}-\cos x& 1\\ -1& -\cos x\end{array}\right|$$

Now, we calculate the individual 2 × 2 determinants:

1. First determinant:
$$\left|\begin{array}{cc}1& \cos x\\ -\cos x& 1\end{array}\right|=1·1-(\cos x)(-\cos x)=1+{\cos }^{2}x$$

2. Second determinant:
$$\left|\begin{array}{cc}-\cos x& \cos x\\ -1& 1\end{array}\right|=-\cos x-(-\cos x)=0$$

3. Third determinant:
$$\left|\begin{array}{cc}-\cos x& 1\\ -1& -\cos x\end{array}\right|={\cos }^{2}x-(-1)={\cos }^{2}x+1$$

Substituting these values back into the expression for $\Delta$:
$$\Delta =1(1+{\cos }^{2}x)-\cos x\left(0\right)+1({\cos }^{2}x+1)$$
$$\Delta =2(1+{\cos }^{2}x)$$

Using the trigonometric identity ${\cos }^{2}x=1-{\sin }^{2}x$:
$$\Delta =2(1+1-{\sin }^{2}x)$$
$$\Delta =2(2-{\sin }^{2}x)$$
This confirms statement (B) is correct, while statement (A) is incorrect.

Evaluating Minimum and Maximum Values:
Since ${\sin }^{2}x$ lies in the range:
$$0\le {\sin }^{2}x\le 1$$

1. Minimum Value of $\Delta$:
The minimum value occurs when ${\sin }^{2}x$ is at its maximum value of 1:
$${\Delta }_{\min }=2(2-1)=2$$
Thus, the minimum value of $\Delta$ is 2, which confirms statement (C) is correct.

2. Maximum Value of $\Delta$:
The maximum value occurs when ${\sin }^{2}x$ is at its minimum value of 0:
$${\Delta }_{\max }=2(2-0)=4$$
Thus, the maximum value of $\Delta$ is 4, which confirms statement (D) is correct.

Consequently, statements (B), (C), and (D) only are correct.