If `tantheta+sintheta=mandtantheta-sintheta=n`,then

Options

A. `m^2-n^2=4mn`
B. `m^2+n^2=4mn`
C. `m^2-n^2=m^2+n^2`
D. `m^2-n^2=4sqrt(mn)`

Explanation

To solve the problem, we are given two equations involving \( \tan \theta \) and \( \sin \theta \):

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

We need to derive a relationship between \( m \), \( n \), and \( \theta \).

### Step 1: Square both equations

First, we will square both equations to express \( m^2 \) and \( n^2 \):

\[ m^2 = (\tan \theta + \sin \theta)^2 = \tan^2 \theta + \sin^2 \theta + 2 \tan \theta \sin \theta \]

\[ n^2 = (\tan \theta - \sin \theta)^2 = \tan^2 \theta + \sin^2 \theta - 2 \tan \theta \sin \theta \]

### Step 2: Subtract the two equations

Now, we will subtract \( n^2 \) from \( m^2 \):

\[ m^2 - n^2 = \left( \tan^2 \theta + \sin^2 \theta + 2 \tan \theta \sin \theta \right) - \left( \tan^2 \theta + \sin^2 \theta - 2 \tan \theta \sin \theta \right) \]

This simplifies to:

\[ m^2 - n^2 = 4 \tan \theta \sin \theta \]

### Step 3: Multiply \( m \) and \( n \)

Next, we will find the product \( mn \):

\[ mn = (\tan \theta + \sin \theta)(\tan \theta - \sin \theta) = \tan^2 \theta - \sin^2 \theta \]

### Step 4: Express \( \tan^2 \theta \) in terms of \( \sin^2 \theta \)

Recall that \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \). Thus, we can rewrite \( mn \):

\[ mn = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta = \sin^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right) = \sin^2 \theta \left( \sec^2 \theta - 1 \right) \]

Using the identity \( \sec^2 \theta - 1 = \tan^2 \theta \):

\[ mn = \sin^2 \theta \tan^2 \theta \]

### Step 5: Relate \( m^2 - n^2 \) to \( mn \)

Now we can express \( 4 \tan \theta \sin \theta \) in terms of \( mn \):

\[ m^2 - n^2 = 4 \tan \theta \sin \theta = 4 \sqrt{mn} \]

### Conclusion

Thus, we have derived the relationship:

\[ m^2 - n^2 = 4 \sqrt{mn} \]