If `x = 1/(2+sqrt3) , y =1/(2-sqrt3)` , then the value of `1/(x+1) +1/(y+1)` is

Options

A. `1/2`
B. `sqrt3`
C. 1
D. `1/sqrt3`

Answer

To solve the problem, we need to find the value of \( \frac{1}{x+1} + \frac{1}{y+1} \) given \( x = \frac{1}{2+\sqrt{3}} \) and \( y = \frac{1}{2-\sqrt{3}} \).

### Step 1: Rationalize \( x \) and \( y \)

First, we will rationalize \( x \) and \( y \).

**For \( x \)**:
\[ x = \frac{1}{2+\sqrt{3}} \cdot \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})} \]
Calculating the denominator:
\[ (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1 \]
Thus,
\[ x = 2 - \sqrt{3} \]

**For \( y \)**:
\[ y = \frac{1}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}} = \frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} \]
Calculating the denominator:
\[ (2-\sqrt{3})(2+\sqrt{3}) = 4 - 3 = 1 \]
Thus,
\[ y = 2 + \sqrt{3} \]

### Step 2: Calculate \( x + 1 \) and \( y + 1 \)

Now we can find \( x + 1 \) and \( y + 1 \):
\[ x + 1 = (2 - \sqrt{3}) + 1 = 3 - \sqrt{3} \]
\[ y + 1 = (2 + \sqrt{3}) + 1 = 3 + \sqrt{3} \]

### Step 3: Calculate \( \frac{1}{x+1} \) and \( \frac{1}{y+1} \)

Next, we compute \( \frac{1}{x+1} \) and \( \frac{1}{y+1} \):
\[ \frac{1}{x+1} = \frac{1}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 + \sqrt{3}}{(3 - \sqrt{3})(3 + \sqrt{3})} \]
Calculating the denominator:
\[ (3 - \sqrt{3})(3 + \sqrt{3}) = 9 - 3 = 6 \]
Thus,
\[ \frac{1}{x+1} = \frac{3 + \sqrt{3}}{6} \]

Similarly for \( y \):
\[ \frac{1}{y+1} = \frac{1}{3 + \sqrt{3}} \cdot \frac{3 - \sqrt{3}}{3 - \sqrt{3}} = \frac{3 - \sqrt{3}}{(3 + \sqrt{3})(3 - \sqrt{3})} \]
Calculating the denominator:
\[ (3 + \sqrt{3})(3 - \sqrt{3}) = 9 - 3 = 6 \]
Thus,
\[ \frac{1}{y+1} = \frac{3 - \sqrt{3}}{6} \]

### Step 4: Combine \( \frac{1}{x+1} + \frac{1}{y+1} \)

Now we can add these two fractions:
\[ \frac{1}{x+1} + \frac{1}{y+1} = \frac{3 + \sqrt{3}}{6} + \frac{3 - \sqrt{3}}{6} = \frac{(3 + \sqrt{3}) + (3 - \sqrt{3})}{6} = \frac{6}{6} = 1 \]

### Final Answer
Thus, the value of \( \frac{1}{x+1} + \frac{1}{y+1} \) is \( \boxed{1} \).