If `cosA=acosB and sinA=bsinB`, then `(b^(2)-a^(2))sin^(2)B=?`

Options

A. `1+a^(2)`
B. `2+a^(2)`
C. `1-a^(2)`
D. `2-a^(2)`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
- \( \cos A = a \cos B \)
- \( \sin A = b \sin B \)

2. **Using the Pythagorean Identity**:
We know that:
\[ \sin^2 A + \cos^2 A = 1 \]
Substituting the values from the given equations:
\[ \sin^2 A = (b \sin B)^2 = b^2 \sin^2 B \]
\[ \cos^2 A = (a \cos B)^2 = a^2 \cos^2 B \]

3. **Substituting into the Identity**:
Now, substituting these into the Pythagorean identity:
\[ b^2 \sin^2 B + a^2 \cos^2 B = 1 \]

4. **Expressing \( \cos^2 B \)**:
We can express \( \cos^2 B \) in terms of \( \sin^2 B \):
\[ \cos^2 B = 1 - \sin^2 B \]
Substituting this into the equation gives:
\[ b^2 \sin^2 B + a^2 (1 - \sin^2 B) = 1 \]

5. **Simplifying the Equation**:
Expanding this:
\[ b^2 \sin^2 B + a^2 - a^2 \sin^2 B = 1 \]
Combining like terms:
\[ (b^2 - a^2) \sin^2 B + a^2 = 1 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ (b^2 - a^2) \sin^2 B = 1 - a^2 \]

7. **Final Result**:
Thus, we have:
\[ (b^2 - a^2) \sin^2 B = 1 - a^2 \]

### Conclusion:
The value of \( (b^2 - a^2) \sin^2 B \) is \( 1 - a^2 \).