The simplification of `((2+sqrt3)/(2-sqrt3)+(2-sqrt3)/(2+sqrt3)+(sqrt3-1)/(sqrt3+1))` is-

Options

A. `2-sqrt3`
B. `2+sqrt3`
C. `16-sqrt3`
D. `40-sqrt3`

Answer

To simplify the expression \(\frac{2+\sqrt{3}}{2-\sqrt{3}} + \frac{2-\sqrt{3}}{2+\sqrt{3}} + \frac{\sqrt{3}-1}{\sqrt{3}+1}\), we can follow these steps:

### Step 1: Find a common denominator for the first two fractions
The common denominator for \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\) and \(\frac{2-\sqrt{3}}{2+\sqrt{3}}\) is \((2-\sqrt{3})(2+\sqrt{3})\).

### Step 2: Rewrite the fractions with the common denominator
\[ \frac{(2+\sqrt{3})^2 + (2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})} \]

### Step 3: Expand the numerators
Using the formula \((a+b)^2 = a^2 + 2ab + b^2\):
- For \((2+\sqrt{3})^2\):
\[ = 2^2 + 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \]
- For \((2-\sqrt{3})^2\):
\[ = 2^2 - 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3} \]

### Step 4: Combine the expanded numerators
\[ (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14 \]

### Step 5: Simplify the denominator
The denominator simplifies as follows:
\[ (2-\sqrt{3})(2+\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \]

### Step 6: Combine the results
Now we have:
\[ \frac{14}{1} = 14 \]

### Step 7: Simplify the last term
Now we need to simplify \(\frac{\sqrt{3}-1}{\sqrt{3}+1}\). We can multiply the numerator and denominator by \(\sqrt{3}-1\):
\[ \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]

### Step 8: Combine all parts
Now we add \(14\) and \(2 - \sqrt{3}\):
\[ 14 + (2 - \sqrt{3}) = 16 - \sqrt{3} \]

### Final Answer
Thus, the simplification of the given expression is:
\[ \boxed{16 - \sqrt{3}} \]