If `costheta+sintheta=m and sectheta+cosectheta=n`, then
What is the value of `n/2(m^2-1)`?

Options

A. m
B. 2m
C. mn
D. 2n

Answer

To solve the problem, we need to find the value of \(\frac{n}{2(m^2 - 1)}\) given that \(m = \cos \theta + \sin \theta\) and \(n = \sec \theta + \csc \theta\).

### Step-by-step Solution:

1. **Express \(m\) in terms of \(\sin\) and \(\cos\)**:
\[ m = \cos \theta + \sin \theta \]

2. **Calculate \(m^2\)**:
\[ m^2 = (\cos \theta + \sin \theta)^2 = \cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta \]
Using the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ m^2 = 1 + 2\sin \theta \cos \theta \]

3. **Find \(m^2 - 1\)**:
\[ m^2 - 1 = 2\sin \theta \cos \theta \]

4. **Express \(n\) in terms of \(\sin\) and \(\cos\)**:
\[ n = \sec \theta + \csc \theta = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \]
Finding a common denominator:
\[ n = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \]
Therefore, substituting \(m\):
\[ n = \frac{m}{\sin \theta \cos \theta} \]

5. **Substituting \(m\) into \(n\)**:
\[ n = \frac{m}{\frac{1}{2} \cdot m^2 - 1} \quad \text{(since } \sin \theta \cos \theta = \frac{1}{2} \cdot m^2 - 1\text{)} \]

6. **Calculate \(\frac{n}{2(m^2 - 1)}\)**:
\[ \frac{n}{2(m^2 - 1)} = \frac{\frac{m}{\sin \theta \cos \theta}}{2(2\sin \theta \cos \theta)} = \frac{m}{4\sin^2 \theta \cos^2 \theta} \]

7. **Simplifying further**:
Since \(\sin^2 \theta + \cos^2 \theta = 1\), we can conclude:
\[ \frac{n}{2(m^2 - 1)} = m \]

### Final Result:
Thus, the value of \(\frac{n}{2(m^2 - 1)}\) is \(m\).