If `sinx+cosx=sqrt7/2` where x`in[0,pi/4]` then `tan(x/2)` is equal to

Options

A. (a) `(3-sqrt(7))/(3)`
B. (b) `(sqrt(7)-2)/(3)`
C. (c) `(4-sqrt(7))/(4)`
D. (d) none of these.

Explanation

To solve the problem, we need to find the value of \( \tan\left(\frac{x}{2}\right) \) given that \( \sin x + \cos x = \frac{\sqrt{7}}{2} \) where \( x \in \left[0, \frac{\pi}{4}\right] \).

### Step 1: Square both sides of the equation
We start with the equation:
\[ \sin x + \cos x = \frac{\sqrt{7}}{2} \]
Squaring both sides gives:
\[ (\sin x + \cos x)^2 = \left(\frac{\sqrt{7}}{2}\right)^2 \]
This simplifies to:
\[ \sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{7}{4} \]

### Step 2: Use the Pythagorean identity
We know from the Pythagorean identity that:
\[ \sin^2 x + \cos^2 x = 1 \]
Substituting this into our equation gives:
\[ 1 + 2 \sin x \cos x = \frac{7}{4} \]
Now, we can solve for \( 2 \sin x \cos x \):
\[ 2 \sin x \cos x = \frac{7}{4} - 1 = \frac{3}{4} \]

### Step 3: Find \( \sin x \cos x \)
Dividing both sides by 2 gives:
\[ \sin x \cos x = \frac{3}{8} \]

### Step 4: Find \( \cos x - \sin x \)
Next, we will find \( \cos x - \sin x \) by squaring it:
\[ (\cos x - \sin x)^2 = \cos^2 x + \sin^2 x - 2 \sin x \cos x \]
Substituting the known values:
\[ (\cos x - \sin x)^2 = 1 - 2 \cdot \frac{3}{8} = 1 - \frac{3}{4} = \frac{1}{4} \]
Taking the square root gives:
\[ \cos x - \sin x = \frac{1}{2} \]
Since \( x \in \left[0, \frac{\pi}{4}\right] \), both \( \cos x \) and \( \sin x \) are positive.

### Step 5: Solve for \( \cos x \) and \( \sin x \)
Now we have two equations:
1. \( \sin x + \cos x = \frac{\sqrt{7}}{2} \)
2. \( \cos x - \sin x = \frac{1}{2} \)

Adding these two equations:
\[ 2\cos x = \frac{\sqrt{7}}{2} + \frac{1}{2} \]
\[ 2\cos x = \frac{\sqrt{7} + 1}{2} \]
Thus,
\[ \cos x = \frac{\sqrt{7} + 1}{4} \]

Now substituting back to find \( \sin x \):
\[ \sin x = \frac{\sqrt{7}}{2} - \cos x = \frac{\sqrt{7}}{2} - \frac{\sqrt{7} + 1}{4} \]
Finding a common denominator:
\[ \sin x = \frac{2\sqrt{7}}{4} - \frac{\sqrt{7} + 1}{4} = \frac{2\sqrt{7} - \sqrt{7} - 1}{4} = \frac{\sqrt{7} - 1}{4} \]

### Step 6: Find \( \tan\left(\frac{x}{2}\right) \)
Using the half-angle formula:
\[ \tan\left(\frac{x}{2}\right) = \frac{1 - \cos x}{\sin x} \]
Substituting the values of \( \cos x \) and \( \sin x \):
\[ \tan\left(\frac{x}{2}\right) = \frac{1 - \frac{\sqrt{7} + 1}{4}}{\frac{\sqrt{7} - 1}{4}} = \frac{\frac{4 - (\sqrt{7} + 1)}{4}}{\frac{\sqrt{7} - 1}{4}} = \frac{4 - \sqrt{7} - 1}{\sqrt{7} - 1} = \frac{3 - \sqrt{7}}{\sqrt{7} - 1} \]

### Step 7: Rationalize the denominator
To rationalize:
\[ \tan\left(\frac{x}{2}\right) = \frac{(3 - \sqrt{7})(\sqrt{7} + 1)}{(\sqrt{7} - 1)(\sqrt{7} + 1)} = \frac{(3\sqrt{7} + 3 - 7 - \sqrt{7})}{7 - 1} = \frac{(2\sqrt{7} - 4)}{6} = \frac{\sqrt{7} - 2}{3} \]

Thus, the final answer is:
\[ \tan\left(\frac{x}{2}\right) = \frac{\sqrt{7} - 2}{3} \]