The value of expression
`(2sin^2 91^@-1)(2sin^2 92^@-1)....(2sin^2 180^@-1)` is equal to

Options

A. `0`
B. `1`
C. `2^(90)`
D. `2^(90)-90`

Answer

To solve the expression \((2\sin^2 91^\circ - 1)(2\sin^2 92^\circ - 1) \ldots (2\sin^2 180^\circ - 1)\), we can follow these steps:

### Step 1: Rewrite each term using the cosine function
We know that:
\[ 2\sin^2 \theta - 1 = -\cos(2\theta) \]
Thus, we can rewrite each term in the product:
\[ 2\sin^2 91^\circ - 1 = -\cos(182^\circ) \]
\[ 2\sin^2 92^\circ - 1 = -\cos(184^\circ) \]
\[ \vdots \]
\[ 2\sin^2 180^\circ - 1 = -\cos(360^\circ) \]

### Step 2: Rewrite the entire expression
The entire expression can be rewritten as:
\[ (-\cos(182^\circ))(-\cos(184^\circ)) \ldots (-\cos(360^\circ)) \]
This can be simplified to:
\[ (-1)^{90} \cos(182^\circ) \cos(184^\circ) \ldots \cos(360^\circ) \]
Since \((-1)^{90} = 1\), we have:
\[ \cos(182^\circ) \cos(184^\circ) \ldots \cos(360^\circ) \]

### Step 3: Identify the value of \(\cos\) at specific angles
Now, we need to evaluate the product of these cosine values. Notably, we have:
- \(\cos(270^\circ) = 0\)

### Step 4: Conclude the value of the product
Since one of the terms in the product is \(\cos(270^\circ) = 0\), the entire product becomes:
\[ \cos(182^\circ) \cos(184^\circ) \ldots \cos(270^\circ) \ldots \cos(360^\circ) = 0 \]

### Final Answer
Thus, the value of the expression \((2\sin^2 91^\circ - 1)(2\sin^2 92^\circ - 1) \ldots (2\sin^2 180^\circ - 1)\) is:
\[ \boxed{0} \]