`(cosx)/(cosy)=n,(sinx)/(siny)=m`, then `(m^(2)-n^(2))sin^(2)y=?`

Options

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

Answer

To solve the problem given by the equations \(\frac{\cos x}{\cos y} = n\) and \(\frac{\sin x}{\sin y} = m\), we need to find the value of \(m^2 - n^2 \sin^2 y\).

### Step-by-step Solution:

1. **Express \(\cos x\) and \(\sin x\) in terms of \(n\) and \(m\)**:
From the first equation, we can express \(\cos x\) as:
\[ \cos x = n \cos y \]
From the second equation, we can express \(\sin x\) as:
\[ \sin x = m \sin y \]

2. **Use the Pythagorean identity**:
We know that:
\[ \sin^2 x + \cos^2 x = 1 \]
Substituting the expressions for \(\sin x\) and \(\cos x\):
\[ (m \sin y)^2 + (n \cos y)^2 = 1 \]
This simplifies to:
\[ m^2 \sin^2 y + n^2 \cos^2 y = 1 \]

3. **Express \(\cos^2 y\) in terms of \(\sin^2 y\)**:
We can use the identity \(\cos^2 y = 1 - \sin^2 y\):
\[ m^2 \sin^2 y + n^2 (1 - \sin^2 y) = 1 \]

4. **Expand and rearrange the equation**:
Expanding the equation gives:
\[ m^2 \sin^2 y + n^2 - n^2 \sin^2 y = 1 \]
Combining like terms results in:
\[ (m^2 - n^2) \sin^2 y + n^2 = 1 \]

5. **Isolate \((m^2 - n^2) \sin^2 y\)**:
Rearranging the equation to isolate \((m^2 - n^2) \sin^2 y\):
\[ (m^2 - n^2) \sin^2 y = 1 - n^2 \]

### Final Result:
Thus, we find that:
\[ (m^2 - n^2) \sin^2 y = 1 - n^2 \]