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

Options

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

Answer

To find the harmonic mean of the two values \(\frac{a}{1 - ab}\) and \(\frac{a}{1 + ab}\), we will use the formula for the harmonic mean of two numbers \(x\) and \(y\):

\[ H = \frac{2xy}{x + y} \]

In our case, we have:

\[ x = \frac{a}{1 - ab}, \quad y = \frac{a}{1 + ab} \]

### Step 1: Calculate \(x + y\)

First, we need to find \(x + y\):

\[ x + y = \frac{a}{1 - ab} + \frac{a}{1 + ab} \]

To add these fractions, we need a common denominator. The common denominator is \((1 - ab)(1 + ab)\).

### Step 2: Find the common denominator and combine

Now, we rewrite the fractions with the common denominator:

\[ x + y = \frac{a(1 + ab)}{(1 - ab)(1 + ab)} + \frac{a(1 - ab)}{(1 - ab)(1 + ab)} \]

Combining the numerators:

\[ x + y = \frac{a(1 + ab) + a(1 - ab)}{(1 - ab)(1 + ab)} \]

### Step 3: Simplify the numerator

Now, simplify the numerator:

\[ x + y = \frac{a(1 + ab + 1 - ab)}{(1 - ab)(1 + ab)} = \frac{a(2)}{(1 - ab)(1 + ab)} = \frac{2a}{(1 - ab)(1 + ab)} \]

### Step 4: Calculate \(xy\)

Next, we calculate \(xy\):

\[ xy = \left(\frac{a}{1 - ab}\right) \left(\frac{a}{1 + ab}\right) = \frac{a^2}{(1 - ab)(1 + ab)} \]

### Step 5: Substitute into the harmonic mean formula

Now we substitute \(x + y\) and \(xy\) into the harmonic mean formula:

\[ H = \frac{2xy}{x + y} = \frac{2 \cdot \frac{a^2}{(1 - ab)(1 + ab)}}{\frac{2a}{(1 - ab)(1 + ab)}} \]

### Step 6: Simplify the expression

This simplifies to:

\[ H = \frac{2a^2}{(1 - ab)(1 + ab)} \cdot \frac{(1 - ab)(1 + ab)}{2a} = \frac{a}{1} \]

Thus, the harmonic mean is:

\[ H = a \]

### Final Answer

The harmonic mean of \(\frac{a}{1 - ab}\) and \(\frac{a}{1 + ab}\) is \(a\).

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