If tan `theta`+ sin `theta` =a and tan `theta`- sin `theta`=b, what is the value `(a^2-b^2) div sqrt(ab)`

Answer

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

1. \( \tan \theta + \sin \theta = a \)
2. \( \tan \theta - \sin \theta = b \)

We need to find the value of \( \frac{a^2 - b^2}{\sqrt{ab}} \).

### Step 1: Use the difference of squares formula
We know that \( a^2 - b^2 = (a - b)(a + b) \). Therefore, we can rewrite our expression as:

\[ \frac{a^2 - b^2}{\sqrt{ab}} = \frac{(a - b)(a + b)}{\sqrt{ab}} \]

### Step 2: Calculate \( a - b \)
Now, let's calculate \( a - b \):

\[ a - b = (\tan \theta + \sin \theta) - (\tan \theta - \sin \theta) = \tan \theta + \sin \theta - \tan \theta + \sin \theta = 2 \sin \theta \]

### Step 3: Calculate \( a + b \)
Next, we calculate \( a + b \):

\[ a + b = (\tan \theta + \sin \theta) + (\tan \theta - \sin \theta) = \tan \theta + \sin \theta + \tan \theta - \sin \theta = 2 \tan \theta \]

### Step 4: Calculate \( ab \)
Now, we calculate \( ab \):

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

Using the identity \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \), we can rewrite \( ab \):

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

Using the Pythagorean identity \( 1 - \cos^2 \theta = \sin^2 \theta \):

\[ ab = \sin^2 \theta \cdot \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\sin^4 \theta}{\cos^2 \theta} \]

### Step 5: Substitute back into the expression
Now substitute \( a - b \), \( a + b \), and \( ab \) into the expression:

\[ \frac{(2 \sin \theta)(2 \tan \theta)}{\sqrt{\frac{\sin^4 \theta}{\cos^2 \theta}}} \]

### Step 6: Simplify the expression
Calculating the square root:

\[ \sqrt{ab} = \sqrt{\frac{\sin^4 \theta}{\cos^2 \theta}} = \frac{\sin^2 \theta}{\cos \theta} \]

Now substituting back:

\[ \frac{(2 \sin \theta)(2 \tan \theta)}{\frac{\sin^2 \theta}{\cos \theta}} = \frac{4 \sin \theta \tan \theta \cos \theta}{\sin^2 \theta} \]

Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):

\[ = \frac{4 \sin \theta \cdot \frac{\sin \theta}{\cos \theta} \cdot \cos \theta}{\sin^2 \theta} = \frac{4 \sin^2 \theta}{\sin^2 \theta} = 4 \]

### Final Answer
Thus, the value of \( \frac{a^2 - b^2}{\sqrt{ab}} \) is:

\[ \boxed{4} \]