If `x = (2ab)/(b^2+1)` then value of `(sqrt(a+x)+sqrt(a-x))/(sqrt(a+x)-sqrt(a-x))` is

Options

A. `a+b`
B. `a-b`
C. a
D. b

Answer

To solve the problem step by step, we start with the given expression and the value of \( x \).

### Step 1: Substitute the value of \( x \)
Given:
\[ x = \frac{2ab}{b^2 + 1} \]
We need to evaluate:
\[ \frac{\sqrt{a+x} + \sqrt{a-x}}{\sqrt{a+x} - \sqrt{a-x}} \]

### Step 2: Calculate \( a + x \) and \( a - x \)
First, we calculate \( a + x \) and \( a - x \):
\[ a + x = a + \frac{2ab}{b^2 + 1} = \frac{a(b^2 + 1) + 2ab}{b^2 + 1} = \frac{ab^2 + a + 2ab}{b^2 + 1} = \frac{ab^2 + 2ab + a}{b^2 + 1} = \frac{a(b^2 + 2b + 1)}{b^2 + 1} = \frac{a(b + 1)^2}{b^2 + 1} \]
\[ a - x = a - \frac{2ab}{b^2 + 1} = \frac{a(b^2 + 1) - 2ab}{b^2 + 1} = \frac{ab^2 + a - 2ab}{b^2 + 1} = \frac{ab^2 - 2ab + a}{b^2 + 1} = \frac{a(b^2 - 2b + 1)}{b^2 + 1} = \frac{a(b - 1)^2}{b^2 + 1} \]

### Step 3: Substitute \( a + x \) and \( a - x \) into the expression
Now substituting these into the original expression:
\[ \frac{\sqrt{a+x} + \sqrt{a-x}}{\sqrt{a+x} - \sqrt{a-x}} = \frac{\sqrt{\frac{a(b + 1)^2}{b^2 + 1}} + \sqrt{\frac{a(b - 1)^2}{b^2 + 1}}}{\sqrt{\frac{a(b + 1)^2}{b^2 + 1}} - \sqrt{\frac{a(b - 1)^2}{b^2 + 1}}} \]

### Step 4: Simplify the expression
Factoring out \( \sqrt{\frac{a}{b^2 + 1}} \):
\[ = \frac{\sqrt{\frac{a}{b^2 + 1}} \left( (b + 1) + (b - 1) \right)}{\sqrt{\frac{a}{b^2 + 1}} \left( (b + 1) - (b - 1) \right)} \]
This simplifies to:
\[ = \frac{\sqrt{\frac{a}{b^2 + 1}} (2b)}{\sqrt{\frac{a}{b^2 + 1}} (2)} = \frac{2b}{2} = b \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{b} \]