If `costheta+sintheta=mandsectheta+"cosec"theta=n` then the value of n `(m^(2)-1)` is equal to :

Options

A. 2 m
B. mn
C. 4mn
D. 2n

Answer

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

### Step 1: Express \( m^2 \)
We start with the equation:
\[ m = \cos \theta + \sin \theta \]
Now, we can square both sides:
\[ m^2 = (\cos \theta + \sin \theta)^2 \]
Using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \), we get:
\[ m^2 = \cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \), we can simplify this to:
\[ m^2 = 1 + 2\cos \theta \sin \theta \]

### Step 2: Find \( m^2 - 1 \)
Now, we subtract 1 from both sides:
\[ m^2 - 1 = 2\cos \theta \sin \theta \]

### Step 3: Express \( n \)
Next, we need to find \( n \):
\[ n = \sec \theta + \csc \theta \]
Using the definitions of secant and cosecant:
\[ n = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \]
Finding a common denominator:
\[ n = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \]
Substituting \( m \) for \( \sin \theta + \cos \theta \):
\[ n = \frac{m}{\sin \theta \cos \theta} \]

### Step 4: Substitute \( \sin \theta \cos \theta \)
From the earlier step, we know:
\[ \sin \theta \cos \theta = \frac{m^2 - 1}{2} \]
Thus, we can substitute this into the expression for \( n \):
\[ n = \frac{m}{\frac{m^2 - 1}{2}} = \frac{2m}{m^2 - 1} \]

### Step 5: Calculate \( n(m^2 - 1) \)
Now we can find \( n(m^2 - 1) \):
\[ n(m^2 - 1) = \left(\frac{2m}{m^2 - 1}\right)(m^2 - 1) \]
The \( m^2 - 1 \) terms cancel out:
\[ n(m^2 - 1) = 2m \]

### Final Answer
Thus, the value of \( n(m^2 - 1) \) is:
\[ \boxed{2m} \]