If `tan theta + sin theta = m` , `tan theta - sin theta = n` then `m^(2) -n^(2)` is equal

Options

A. `sqrt(mn)`
B. `sqrt(m/n)`
C. `4sqrt(mn)`
D. none of these

Explanation

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

1. \( m = \tan \theta + \sin \theta \)
2. \( n = \tan \theta - \sin \theta \)

We need to find \( m^2 - n^2 \).

### Step 1: Use the difference of squares formula
We can use the difference of squares formula, which states that \( a^2 - b^2 = (a - b)(a + b) \).

So, we have:
\[ m^2 - n^2 = (m - n)(m + n) \]

### Step 2: Calculate \( m - n \) and \( m + n \)
First, we calculate \( m - n \):
\[ m - n = (\tan \theta + \sin \theta) - (\tan \theta - \sin \theta) = 2 \sin \theta \]

Next, we calculate \( m + n \):
\[ m + n = (\tan \theta + \sin \theta) + (\tan \theta - \sin \theta) = 2 \tan \theta \]

### Step 3: Substitute back into the difference of squares formula
Now we substitute \( m - n \) and \( m + n \) back into the formula:
\[ m^2 - n^2 = (2 \sin \theta)(2 \tan \theta) = 4 \sin \theta \tan \theta \]

### Step 4: Simplify \( \sin \theta \tan \theta \)
Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Therefore:
\[ \sin \theta \tan \theta = \sin \theta \cdot \frac{\sin \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \]

### Step 5: Final expression for \( m^2 - n^2 \)
Substituting this back into our expression gives:
\[ m^2 - n^2 = 4 \cdot \frac{\sin^2 \theta}{\cos \theta} \]

Thus, the final answer is:
\[ m^2 - n^2 = \frac{4 \sin^2 \theta}{\cos \theta} \]