The harmonic mean of `(a)/(1-ab) and (a)/(1 + ab)` is

Options

A. `(a)/(sqrt((1 - a^(2) b^(2))))`
B. `(a)/(1 - a^(2)b^(2))`
C. a
D. `(1)/(1 - a^(2) b^(2))`

Answer

To find the harmonic mean of the two numbers \(\frac{a}{1-ab}\) and \(\frac{a}{1+ab}\), we can follow these steps:

### Step 1: Identify the formula for the harmonic mean
The harmonic mean (HM) of two numbers \(P\) and \(Q\) is given by the formula:
\[ HM = \frac{2PQ}{P + Q} \]

### Step 2: Assign values to \(P\) and \(Q\)
Let:
\[ P = \frac{a}{1-ab} \quad \text{and} \quad Q = \frac{a}{1+ab} \]

### Step 3: Calculate \(P + Q\)
To find \(P + Q\), we need a common denominator:
\[ P + Q = \frac{a}{1-ab} + \frac{a}{1+ab} = \frac{a(1+ab) + a(1-ab)}{(1-ab)(1+ab)} \]
Simplifying the numerator:
\[ = \frac{a(1 + ab + 1 - ab)}{(1-ab)(1+ab)} = \frac{a(2)}{(1-ab)(1+ab)} = \frac{2a}{(1-ab)(1+ab)} \]

### Step 4: Calculate \(PQ\)
Now, we calculate \(PQ\):
\[ PQ = \left(\frac{a}{1-ab}\right) \left(\frac{a}{1+ab}\right) = \frac{a^2}{(1-ab)(1+ab)} \]

### Step 5: Substitute \(P + Q\) and \(PQ\) into the HM formula
Now we substitute \(P + Q\) and \(PQ\) into the harmonic mean formula:
\[ HM = \frac{2PQ}{P + Q} = \frac{2 \cdot \frac{a^2}{(1-ab)(1+ab)}}{\frac{2a}{(1-ab)(1+ab)}} \]

### Step 6: Simplify the expression
The \((1-ab)(1+ab)\) cancels out:
\[ HM = \frac{2a^2}{2a} = a \]

### Conclusion
Thus, the harmonic mean of \(\frac{a}{1-ab}\) and \(\frac{a}{1+ab}\) is:
\[ \boxed{a} \]