The value of expression `(2sin^2 91^0-1)(2sin^2 92^0-1)...(2sin^2 180^0-1)` is equal to 0 (b) `1` (c) `2^(90)` (d) `2^(90)-90`

Options

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

Explanation

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

### Step 1: Identify the terms in the product
The expression consists of terms from \(2\sin^2 91^\circ - 1\) to \(2\sin^2 180^\circ - 1\). We can express this as:
\[ \prod_{k=91}^{180} (2\sin^2 k^\circ - 1) \]

### Step 2: Simplify the individual terms
Recall the identity:
\[ 2\sin^2 \theta - 1 = \cos(2\theta) \]
Using this identity, we can rewrite each term:
\[ 2\sin^2 k^\circ - 1 = \cos(2k^\circ) \]
Thus, the product becomes:
\[ \prod_{k=91}^{180} \cos(2k^\circ) \]

### Step 3: Evaluate specific terms
Let's evaluate the middle term when \(k = 135^\circ\):
\[ 2\sin^2 135^\circ - 1 = 2\left(\frac{1}{\sqrt{2}}\right)^2 - 1 = 2 \cdot \frac{1}{2} - 1 = 0 \]
This shows that one of the terms in the product is zero.

### Step 4: Conclude the product
Since one of the terms in the product is zero, the entire product evaluates to zero:
\[ (2\sin^2 91^\circ - 1)(2\sin^2 92^\circ - 1)\cdots(2\sin^2 180^\circ - 1) = 0 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{0} \]