If `tantheta+sintheta=mandtantheta-sintheta=n`, then find the value of `sqrt(mn)`.

Options

A. `(1)/(2)(m^(2)-n^(2))`
B. `2(m^(2)-n^(2))`
C. `(1)/(4)(m^(2)-n^(2))`
D. `(1)/(4)(m^(2)+n^(2))`

Answer

To solve the problem where \( \tan \theta + \sin \theta = m \) and \( \tan \theta - \sin \theta = n \), we need to find the value of \( \sqrt{mn} \).

### Step-by-Step Solution:

1. **Square Both Equations**:
- From the first equation:
\[ (\tan \theta + \sin \theta)^2 = m^2 \]
Expanding this gives:
\[ \tan^2 \theta + 2 \tan \theta \sin \theta + \sin^2 \theta = m^2 \tag{1} \]

- From the second equation:
\[ (\tan \theta - \sin \theta)^2 = n^2 \]
Expanding this gives:
\[ \tan^2 \theta - 2 \tan \theta \sin \theta + \sin^2 \theta = n^2 \tag{2} \]

2. **Subtract the Two Equations**:
- Subtract equation (2) from equation (1):
\[ [\tan^2 \theta + 2 \tan \theta \sin \theta + \sin^2 \theta] - [\tan^2 \theta - 2 \tan \theta \sin \theta + \sin^2 \theta] = m^2 - n^2 \]
- This simplifies to:
\[ 4 \tan \theta \sin \theta = m^2 - n^2 \]

3. **Express \( \tan \theta \) in terms of \( \sin \theta \)**:
- Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ 4 \left(\frac{\sin \theta}{\cos \theta}\right) \sin \theta = m^2 - n^2 \]
- This can be rewritten as:
\[ 4 \frac{\sin^2 \theta}{\cos \theta} = m^2 - n^2 \]

4. **Rearranging the Equation**:
- Multiply both sides by \( \cos \theta \):
\[ 4 \sin^2 \theta = (m^2 - n^2) \cos \theta \]

5. **Finding \( \sqrt{mn} \)**:
- From the earlier equation \( 4 \tan \theta \sin \theta = m^2 - n^2 \), we can express \( mn \) as:
\[ mn = \left(\tan \theta + \sin \theta\right)\left(\tan \theta - \sin \theta\right) = \tan^2 \theta - \sin^2 \theta \]
- Using \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \):
\[ mn = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta = \sin^2 \theta \left(\frac{1 - \cos^2 \theta}{\cos^2 \theta}\right) = \frac{\sin^2 \theta \sin^2 \theta}{\cos^2 \theta} = \frac{\sin^4 \theta}{\cos^2 \theta} \]

6. **Taking the Square Root**:
- Now we take the square root:
\[ \sqrt{mn} = \sqrt{\frac{\sin^4 \theta}{\cos^2 \theta}} = \frac{\sin^2 \theta}{\cos \theta} = \tan \theta \sin \theta \]

### Final Answer:
Thus, the value of \( \sqrt{mn} \) is \( \tan \theta \sin \theta \).