If `cosectheta-sintheta=mandsectheta-costheta=n`, then `(m^(2)n)^((2)/(3))+(mn^(2))^((2)/(3))=?`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the values of \( m \) and \( n \) based on the given equations and then compute the expression \( (m^2 n)^{\frac{2}{3}} + (m n^2)^{\frac{2}{3}} \).

### Step-by-step Solution:

1. **Finding \( m \)**:
Given \( m = \csc \theta - \sin \theta \).

We know that:
\[ \csc \theta = \frac{1}{\sin \theta} \]
Therefore,
\[ m = \frac{1}{\sin \theta} - \sin \theta = \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta} \]

2. **Finding \( n \)**:
Given \( n = \sec \theta - \cos \theta \).

We know that:
\[ \sec \theta = \frac{1}{\cos \theta} \]
Therefore,
\[ n = \frac{1}{\cos \theta} - \cos \theta = \frac{1 - \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \]

3. **Calculating \( m^2 n \)**:
Now we compute \( m^2 n \):
\[ m^2 = \left(\frac{\cos^2 \theta}{\sin \theta}\right)^2 = \frac{\cos^4 \theta}{\sin^2 \theta} \]
Therefore,
\[ m^2 n = \frac{\cos^4 \theta}{\sin^2 \theta} \cdot \frac{\sin^2 \theta}{\cos \theta} = \frac{\cos^4 \theta}{\cos \theta} = \cos^3 \theta \]

4. **Calculating \( mn^2 \)**:
Now we compute \( mn^2 \):
\[ n^2 = \left(\frac{\sin^2 \theta}{\cos \theta}\right)^2 = \frac{\sin^4 \theta}{\cos^2 \theta} \]
Therefore,
\[ mn^2 = \frac{\cos^2 \theta}{\sin \theta} \cdot \frac{\sin^4 \theta}{\cos^2 \theta} = \frac{\sin^4 \theta}{\sin \theta} = \sin^3 \theta \]

5. **Calculating the final expression**:
Now we compute:
\[ (m^2 n)^{\frac{2}{3}} + (mn^2)^{\frac{2}{3}} = (\cos^3 \theta)^{\frac{2}{3}} + (\sin^3 \theta)^{\frac{2}{3}} \]
This simplifies to:
\[ \cos^2 \theta + \sin^2 \theta \]
We know that:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]

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