If `tanA=ntanBandsinA=msinB`then `sec^(3)A((m^(2)-1)/(n^(2)-1))^((3)/(2)`=?

Options

A. 1
B. `cos^(2)A`
C. 0
D. `cos^(3)A`

Answer

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

1. **Given Equations**:
\[ \tan A = n \tan B \quad \text{and} \quad \sin A = m \sin B \]

2. **Convert Tangent and Sine**:
From the first equation, we can express \(\tan B\) in terms of \(\tan A\):
\[ \tan B = \frac{\tan A}{n} \]
From the second equation, we can express \(\sin B\) in terms of \(\sin A\):
\[ \sin B = \frac{\sin A}{m} \]

3. **Using Cotangent and Cosecant**:
We know that:
\[ \cot B = \frac{1}{\tan B} = \frac{n \cos A}{\sin A} \]
and
\[ \csc B = \frac{1}{\sin B} = \frac{m}{\sin A} \]

4. **Using the Identity**:
We use the identity:
\[ \cos^2 B - \cot^2 B = 1 \]
Substituting the values of \(\cos B\) and \(\cot B\):
\[ \cos^2 B - \left(\frac{n \cos A}{\sin A}\right)^2 = 1 \]

5. **Substituting Values**:
We can express \(\cos^2 B\) in terms of \(\sin A\) and \(\cos A\):
\[ \cos^2 B = \frac{m^2}{\sin^2 A} \]
Therefore, substituting this into the identity gives:
\[ \frac{m^2}{\sin^2 A} - \frac{n^2 \cos^2 A}{\sin^2 A} = 1 \]

6. **Finding a Common Denominator**:
Combine the fractions:
\[ \frac{m^2 - n^2 \cos^2 A}{\sin^2 A} = 1 \]
This implies:
\[ m^2 - n^2 \cos^2 A = \sin^2 A \]

7. **Using the Pythagorean Identity**:
We can use the identity \(\sin^2 A = 1 - \cos^2 A\):
\[ m^2 - n^2 \cos^2 A = 1 - \cos^2 A \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ m^2 - 1 = n^2 \cos^2 A - \cos^2 A \]
This simplifies to:
\[ m^2 - 1 = (n^2 - 1) \cos^2 A \]

9. **Solving for \(\cos^2 A\)**:
Thus, we can express \(\cos^2 A\) as:
\[ \cos^2 A = \frac{m^2 - 1}{n^2 - 1} \]

10. **Finding the Required Expression**:
Now we need to find:
\[ \sec^3 A \left(\frac{m^2 - 1}{n^2 - 1}\right)^{\frac{3}{2}} \]
Using \(\sec A = \frac{1}{\cos A}\), we have:
\[ \sec^3 A = \frac{1}{\cos^3 A} \]
Therefore:
\[ \sec^3 A = \frac{1}{\left(\frac{m^2 - 1}{n^2 - 1}\right)^{\frac{3}{2}}} \]

11. **Final Calculation**:
Substituting this into our expression gives:
\[ \sec^3 A \left(\frac{m^2 - 1}{n^2 - 1}\right)^{\frac{3}{2}} = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]