If `(cosA)/(cosB)=n and (sinA)/(sinB)=m,then (m^(2)-n^(2))sin^(2)B=`

Options

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

Answer

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

1. \(\frac{\cos A}{\cos B} = n\)
2. \(\frac{\sin A}{\sin B} = m\)

We need to find the value of the expression \( (m^2 - n^2) \sin^2 B \).

### Step 1: Express \(m^2\) and \(n^2\)

From the definitions of \(m\) and \(n\):
- \(m = \frac{\sin A}{\sin B}\) implies \(m^2 = \frac{\sin^2 A}{\sin^2 B}\)
- \(n = \frac{\cos A}{\cos B}\) implies \(n^2 = \frac{\cos^2 A}{\cos^2 B}\)

### Step 2: Substitute \(m^2\) and \(n^2\) into the expression

Now, substitute \(m^2\) and \(n^2\) into the expression \( (m^2 - n^2) \sin^2 B \):

\[ (m^2 - n^2) \sin^2 B = \left(\frac{\sin^2 A}{\sin^2 B} - \frac{\cos^2 A}{\cos^2 B}\right) \sin^2 B \]

### Step 3: Simplify the expression

Distributing \(\sin^2 B\) gives:

\[ = \sin^2 A \cdot \frac{\sin^2 B}{\sin^2 B} - \cos^2 A \cdot \frac{\sin^2 B}{\cos^2 B} \]
\[ = \sin^2 A - \cos^2 A \cdot \frac{\sin^2 B}{\cos^2 B} \]

### Step 4: Rewrite \(\sin^2 A\) and \(\sin^2 B\)

Using the Pythagorean identity, we know that:

\[ \sin^2 A = 1 - \cos^2 A \]

Substituting this into the expression gives:

\[ = (1 - \cos^2 A) - \cos^2 A \cdot \frac{\sin^2 B}{\cos^2 B} \]

### Step 5: Combine the terms

Now, we can combine the terms:

\[ = 1 - \cos^2 A - \cos^2 A \cdot \frac{\sin^2 B}{\cos^2 B} \]
\[ = 1 - \cos^2 A \left(1 + \frac{\sin^2 B}{\cos^2 B}\right) \]

### Step 6: Simplify further

Recall that \(\frac{\sin^2 B}{\cos^2 B} = \tan^2 B\), so we can rewrite:

\[ = 1 - \cos^2 A (1 + \tan^2 B) \]

Using the identity \(1 + \tan^2 B = \sec^2 B\):

\[ = 1 - \cos^2 A \sec^2 B \]

### Step 7: Final expression

Since \(\sec^2 B = \frac{1}{\cos^2 B}\), we can rewrite the expression as:

\[ = 1 - \frac{\cos^2 A}{\cos^2 B} \]

### Step 8: Relate back to \(n^2\)

Recall that \(n^2 = \frac{\cos^2 A}{\cos^2 B}\), thus:

\[ = 1 - n^2 \]

### Conclusion

The final result is:

\[ (m^2 - n^2) \sin^2 B = 1 - n^2 \]