If `alpha + beta + gamma=pi, ` then the value of `sin ^(2) alpha + sin ^(2) beta - sin^(2) gamma `is equal to

Options

A. `2 sin alpha `
B. `2 sin alpha cos beta sin gamma`
C. `2 sin alpha sin beta cos gamma`
D. `2 sin alpha sin beta sin gamma`

Answer

To find the value of \( \sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma \) given that \( \alpha + \beta + \gamma = \pi \), we can follow these steps:

### Step 1: Use the identity for sine
We know that \( \sin^2 \gamma = \sin^2(\pi - (\alpha + \beta)) \). Using the identity \( \sin(\pi - x) = \sin x \), we have:
\[ \sin^2 \gamma = \sin^2(\alpha + \beta) \]

### Step 2: Expand \( \sin^2(\alpha + \beta) \)
Using the sine addition formula:
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]
Thus,
\[ \sin^2(\alpha + \beta) = (\sin \alpha \cos \beta + \cos \alpha \sin \beta)^2 \]
Expanding this, we get:
\[ \sin^2(\alpha + \beta) = \sin^2 \alpha \cos^2 \beta + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta + \cos^2 \alpha \sin^2 \beta \]

### Step 3: Substitute back into the expression
Now substituting this back into our original expression:
\[ \sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta) \]
This becomes:
\[ \sin^2 \alpha + \sin^2 \beta - (\sin^2 \alpha \cos^2 \beta + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta + \cos^2 \alpha \sin^2 \beta) \]

### Step 4: Simplify the expression
Combining like terms, we have:
\[ \sin^2 \alpha + \sin^2 \beta - \sin^2 \alpha \cos^2 \beta - 2 \sin \alpha \cos \beta \cos \alpha \sin \beta - \cos^2 \alpha \sin^2 \beta \]
This can be rearranged to:
\[ \sin^2 \alpha (1 - \cos^2 \beta) + \sin^2 \beta (1 - \cos^2 \alpha) - 2 \sin \alpha \cos \beta \cos \alpha \sin \beta \]
Using the identity \( 1 - \cos^2 x = \sin^2 x \):
\[ \sin^2 \alpha \sin^2 \beta + \sin^2 \beta \sin^2 \alpha - 2 \sin \alpha \cos \beta \cos \alpha \sin \beta \]

### Step 5: Final result
This simplifies to:
\[ \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta) = \sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = \sin^2 \alpha + \sin^2 \beta - \sin^2(\pi - (\alpha + \beta)) \]
Thus, the final answer is:
\[ \sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = 1 \]