If `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, we need to find the value of \( \tan\left(\frac{A}{2}\right) \) given that \( A < \frac{\pi}{6} \) and \( \sin A + \cos A = \frac{\sqrt{7}}{2} \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \sin A + \cos A = \frac{\sqrt{7}}{2} \]

2. **Use the identities for sine and cosine in terms of \( \tan\left(\frac{A}{2}\right) \):**
\[ \sin A = \frac{2\tan\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)} \quad \text{and} \quad \cos A = \frac{1 - \tan^2\left(\frac{A}{2}\right)}{1 + \tan^2\left(\frac{A}{2}\right)} \]
Let \( x = \tan\left(\frac{A}{2}\right) \).

3. **Substituting the identities into the equation:**
\[ \frac{2x}{1 + x^2} + \frac{1 - x^2}{1 + x^2} = \frac{\sqrt{7}}{2} \]

4. **Combine the fractions:**
\[ \frac{2x + 1 - x^2}{1 + x^2} = \frac{\sqrt{7}}{2} \]

5. **Cross-multiply to eliminate the fraction:**
\[ 2(2x + 1 - x^2) = \sqrt{7}(1 + x^2) \]
Simplifying gives:
\[ 4x + 2 - 2x^2 = \sqrt{7} + \sqrt{7}x^2 \]

6. **Rearranging the equation:**
\[ (2 + \sqrt{7})x^2 - 4x + (2 - \sqrt{7}) = 0 \]

7. **Using the quadratic formula to solve for \( x \):**
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2 + \sqrt{7} \), \( b = -4 \), and \( c = 2 - \sqrt{7} \).

8. **Calculating the discriminant:**
\[ b^2 - 4ac = (-4)^2 - 4(2 + \sqrt{7})(2 - \sqrt{7}) = 16 - 4[(2)(2) - (7)] = 16 - 4(4 - 7) = 16 - 4(-3) = 16 + 12 = 28 \]

9. **Substituting back into the quadratic formula:**
\[ x = \frac{4 \pm \sqrt{28}}{2(2 + \sqrt{7})} = \frac{4 \pm 2\sqrt{7}}{4 + 2\sqrt{7}} \]

10. **Simplifying the expression:**
\[ x = \frac{2 \pm \sqrt{7}}{2 + \sqrt{7}} \]

11. **Choosing the appropriate solution:**
Since \( A < \frac{\pi}{6} \), \( \tan\left(\frac{A}{2}\right) \) must be positive and less than 1. Thus, we take:
\[ x = \frac{2 - \sqrt{7}}{2 + \sqrt{7}} \]

12. **Rationalizing the denominator:**
\[ x = \frac{(2 - \sqrt{7})(2 - \sqrt{7})}{(2 + \sqrt{7})(2 - \sqrt{7})} = \frac{(2 - \sqrt{7})^2}{4 - 7} = \frac{4 - 4\sqrt{7} + 7}{-3} = \frac{11 - 4\sqrt{7}}{-3} \]

### Final Result:
Thus, the value of \( \tan\left(\frac{A}{2}\right) \) is:
\[ \tan\left(\frac{A}{2}\right) = \frac{2 - \sqrt{7}}{2 + \sqrt{7}} \]