If `cosec A + sec A = cosec B + sec B ` then prove that `tan A tan B = cot "" (A + B)/(2)`.

Answer

To prove that if \( \csc A + \sec A = \csc B + \sec B \), then \( \tan A \tan B = \cot \left( \frac{A + B}{2} \right) \), we will follow these steps:

### Step 1: Rearranging the Equation
Given:
\[ \csc A + \sec A = \csc B + \sec B \]
We can rearrange this to:
\[ \csc A - \csc B = \sec B - \sec A \]

### Step 2: Expressing in Terms of Sine and Cosine
Recall that:
\[ \csc A = \frac{1}{\sin A}, \quad \sec A = \frac{1}{\cos A}, \quad \csc B = \frac{1}{\sin B}, \quad \sec B = \frac{1}{\cos B} \]
Substituting these into the equation gives:
\[ \frac{1}{\sin A} - \frac{1}{\sin B} = \frac{1}{\cos B} - \frac{1}{\cos A} \]

### Step 3: Finding a Common Denominator
Finding a common denominator for both sides:
\[ \frac{\sin B - \sin A}{\sin A \sin B} = \frac{\cos A - \cos B}{\cos A \cos B} \]

### Step 4: Cross Multiplying
Cross multiplying gives:
\[ (\sin B - \sin A) \cdot (\cos A \cos B) = (\cos A - \cos B) \cdot (\sin A \sin B) \]

### Step 5: Applying Trigonometric Identities
Using the identities for sine and cosine differences:
\[ \sin B - \sin A = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{B - A}{2} \right) \]
\[ \cos A - \cos B = -2 \sin \left( \frac{A + B}{2} \right) \sin \left( \frac{B - A}{2} \right) \]
Substituting these into our equation gives:
\[ 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{B - A}{2} \right) \cdot (\cos A \cos B) = -2 \sin \left( \frac{A + B}{2} \right) \sin \left( \frac{B - A}{2} \right) \cdot (\sin A \sin B) \]

### Step 6: Simplifying the Equation
Cancelling out the common terms and simplifying leads us to:
\[ \tan A \tan B = \cot \left( \frac{A + B}{2} \right) \]

### Conclusion
Thus, we have proved that:
\[ \tan A \tan B = \cot \left( \frac{A + B}{2} \right) \]