If `sinA sinB sinC+cosAcosB=1,` then the value of sinC is

Options

A. 1
B. `1//2`
C. 0
D. `-1`

Explanation

To solve the equation \( \sin A \sin B \sin C + \cos A \cos B = 1 \) and find the value of \( \sin C \), we can follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ \sin A \sin B \sin C + \cos A \cos B = 1 \]

### Step 2: Isolate the sine term
Rearranging the equation gives:
\[ \sin A \sin B \sin C = 1 - \cos A \cos B \]

### Step 3: Use the Pythagorean identity
Recall that \( \sin^2 A + \cos^2 A = 1 \) and \( \sin^2 B + \cos^2 B = 1 \). This implies:
\[ \cos A \cos B = \sqrt{(1 - \sin^2 A)(1 - \sin^2 B)} \]
However, for simplicity, we will not expand this further and will keep it in terms of \( \cos A \) and \( \cos B \).

### Step 4: Multiply the entire equation by 2
Multiply the entire equation by 2:
\[ 2 \sin A \sin B \sin C + 2 \cos A \cos B = 2 \]

### Step 5: Recognize the identity
Using the identity \( \sin^2 A + \cos^2 A = 1 \) and \( \sin^2 B + \cos^2 B = 1 \), we can express:
\[ 2 \sin A \sin B \sin C + 2 \cos A \cos B = \sin^2 A + \cos^2 A + \sin^2 B + \cos^2 B = 2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 2 \sin A \sin B \sin C + 2 \cos A \cos B - 2 = 0 \]

### Step 7: Factor the equation
We can factor the equation:
\[ (\sin A \sin B)^2 + (\cos A - \cos B)^2 + 2 \sin A \sin B (1 - \sin C) = 0 \]

### Step 8: Set each term to zero
For the equation to hold true, each term must equal zero:
1. \( \sin A \sin B = 0 \)
2. \( \cos A - \cos B = 0 \)
3. \( 1 - \sin C = 0 \)

### Step 9: Solve for \( \sin C \)
From the third condition:
\[ 1 - \sin C = 0 \implies \sin C = 1 \]

### Conclusion
Thus, the value of \( \sin C \) is:
\[ \sin C = 1 \]