If 0`ltA lt(pi)/(6) and sinA+cosA=(sqrt7)/(2),"then" tan""(A)/(2)=`

Options

A. `(sqrt7-2)/(3)`
B. `(sqrt7+2)/(3)`
C. `(sqrt7)/(3)`
D. none of these

Answer

To solve the problem step-by-step, we need to find the value of \( \tan\left(\frac{A}{2}\right) \) given that \( \sin A + \cos A = \frac{\sqrt{7}}{2} \) and \( 0 < A < \frac{\pi}{6} \).

### Step 1: Write the given equation
We start with the equation:
\[ \sin A + \cos A = \frac{\sqrt{7}}{2} \]

### Step 2: Use the identity for sine and cosine
We can express \( \sin A \) and \( \cos A \) in terms of \( \tan\left(\frac{A}{2}\right) \):
\[ \sin A = \frac{2 \tan\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)} \]
\[ \cos A = \frac{1 - \tan^2\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)} \]

### Step 3: Substitute into the equation
Substituting these identities into the equation gives:
\[ \frac{2 \tan\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)} + \frac{1 - \tan^2\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)} = \frac{\sqrt{7}}{2} \]

### Step 4: Combine the fractions
Combining the fractions on the left side:
\[ \frac{2 \tan\left(\frac{A}{2}\right) + 1 - \tan^2\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)} = \frac{\sqrt{7}}{2} \]

### Step 5: Cross-multiply
Cross-multiplying gives:
\[ 2(2 \tan\left(\frac{A}{2}\right) + 1 - \tan^2\left(\frac{A}{2}\right)) = \sqrt{7}(1 + \tan^2\left(\frac{A}{2}\right)) \]

### Step 6: Rearranging the equation
Expanding and rearranging:
\[ 4 \tan\left(\frac{A}{2}\right) + 2 - 2 \tan^2\left(\frac{A}{2}\right) = \sqrt{7} + \sqrt{7} \tan^2\left(\frac{A}{2}\right) \]
\[ (2 + \sqrt{7}) \tan^2\left(\frac{A}{2}\right) - 4 \tan\left(\frac{A}{2}\right) + (\sqrt{7} - 2) = 0 \]

### Step 7: Identify coefficients for the quadratic formula
Let \( x = \tan\left(\frac{A}{2}\right) \). The equation becomes:
\[ (2 + \sqrt{7})x^2 - 4x + (\sqrt{7} - 2) = 0 \]
Here, \( a = 2 + \sqrt{7} \), \( b = -4 \), and \( c = \sqrt{7} - 2 \).

### Step 8: Use the quadratic formula
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values:
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4(2 + \sqrt{7})(\sqrt{7} - 2)}}{2(2 + \sqrt{7})} \]
Calculating the discriminant:
\[ 16 - 4(2 + \sqrt{7})(\sqrt{7} - 2) \]

### Step 9: Simplify the discriminant
Calculating:
\[ = 16 - 4[(2\sqrt{7} - 4) + (7 - 2\sqrt{7})] = 16 - 4(3) = 4 \]
So,
\[ x = \frac{4 \pm 2}{2(2 + \sqrt{7})} \]

### Step 10: Find the two possible values
Calculating the two values:
1. \( x = \frac{6}{2(2 + \sqrt{7})} = \frac{3}{2 + \sqrt{7}} \)
2. \( x = \frac{2}{2(2 + \sqrt{7})} = \frac{1}{2 + \sqrt{7}} \)

### Step 11: Rationalize the denominators
For \( x = \frac{3}{2 + \sqrt{7}} \):
\[ = \frac{3(2 - \sqrt{7})}{(2 + \sqrt{7})(2 - \sqrt{7})} = \frac{6 - 3\sqrt{7}}{-3} = \sqrt{7} - 2 \]

For \( x = \frac{1}{2 + \sqrt{7}} \):
\[ = \frac{(2 - \sqrt{7})}{(2 + \sqrt{7})(2 - \sqrt{7})} = \frac{2 - \sqrt{7}}{-3} \]

### Final Result
Thus, the possible values for \( \tan\left(\frac{A}{2}\right) \) are:
1. \( \tan\left(\frac{A}{2}\right) = \sqrt{7} - 2 \)
2. \( \tan\left(\frac{A}{2}\right) = \frac{2 - \sqrt{7}}{3} \)

Since \( 0 < A < \frac{\pi}{6} \), we take the positive value:
\[ \tan\left(\frac{A}{2}\right) = \frac{\sqrt{7} - 2}{3} \]