If `a=(sqrt(3))/(2)` , then the value of `sqrt(1+a)+sqrt(1-a)` is

Options

A. `sqrt(3)`
B. `(sqrt(3))/(2)`
C. `2+sqrt(3)`
D. `2-sqrt(3)`

Answer

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

### Step-by-Step Solution:

**Step 1: Substitute the value of \( a \)**
We start by substituting \( a \) into the expression:
\[ \sqrt{1 + a} + \sqrt{1 - a} = \sqrt{1 + \frac{\sqrt{3}}{2}} + \sqrt{1 - \frac{\sqrt{3}}{2}} \]

**Step 2: Simplify \( 1 + a \) and \( 1 - a \)**
Calculate \( 1 + \frac{\sqrt{3}}{2} \) and \( 1 - \frac{\sqrt{3}}{2} \):
\[ 1 + \frac{\sqrt{3}}{2} = \frac{2}{2} + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2} \]
\[ 1 - \frac{\sqrt{3}}{2} = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \]

**Step 3: Rewrite the expression**
Now we can rewrite the expression:
\[ \sqrt{1 + a} + \sqrt{1 - a} = \sqrt{\frac{2 + \sqrt{3}}{2}} + \sqrt{\frac{2 - \sqrt{3}}{2}} \]

**Step 4: Factor out \( \sqrt{2} \)**
Factor out \( \sqrt{2} \) from both terms:
\[ = \sqrt{2} \left( \sqrt{\frac{2 + \sqrt{3}}{4}} + \sqrt{\frac{2 - \sqrt{3}}{4}} \right) \]
\[ = \sqrt{2} \left( \frac{\sqrt{2 + \sqrt{3}}}{2} + \frac{\sqrt{2 - \sqrt{3}}}{2} \right) \]

**Step 5: Combine the terms**
Combine the terms inside the parentheses:
\[ = \frac{\sqrt{2}}{2} \left( \sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}} \right) \]

**Step 6: Simplify further**
Now we need to simplify \( \sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}} \). We can use the identity:
\[ \sqrt{a} + \sqrt{b} = \sqrt{(a+b) + 2\sqrt{ab}} \]
Let \( a = 2 + \sqrt{3} \) and \( b = 2 - \sqrt{3} \):
\[ a + b = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \]
\[ ab = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1 \]
Thus,
\[ \sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}} = \sqrt{4 + 2\sqrt{1}} = \sqrt{4 + 2} = \sqrt{6} \]

**Step 7: Final expression**
Substituting back, we have:
\[ \sqrt{1 + a} + \sqrt{1 - a} = \frac{\sqrt{2}}{2} \cdot \sqrt{6} = \sqrt{3} \]

### Final Answer:
The value of \( \sqrt{1+a} + \sqrt{1-a} \) is \( \sqrt{3} \).