In `triangleABC, tan((A)/(2))tan((B)/(2))=`

Options

A. `(c-(a+b))/(2(a+b+c))`
B. `(c-(a+b))/(a+b+c)`
C. `(a+b-c)/(2(a+b+c))`
D. `(a+b-c)/(a+b+c)`

Answer

To solve the problem, we need to find the expression for \( \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) \) in triangle \( ABC \).

### Step-by-Step Solution:

1. **Use the Half-Angle Formula for Tangent**:
The half-angle formulas for the tangent of angles \( A \) and \( B \) in triangle \( ABC \) are given by:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \]
\[ \tan\left(\frac{B}{2}\right) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \]
where \( s \) is the semi-perimeter of the triangle, defined as:
\[ s = \frac{a + b + c}{2} \]

2. **Multiply the Two Expressions**:
Now, we multiply \( \tan\left(\frac{A}{2}\right) \) and \( \tan\left(\frac{B}{2}\right) \):
\[ \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \cdot \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \]

3. **Combine the Square Roots**:
Combining the square roots gives:
\[ \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) = \sqrt{\frac{(s-b)(s-c)(s-a)(s-c)}{s^2(s-a)(s-b)}} \]

4. **Simplify the Expression**:
Notice that \( (s-c) \) appears in both the numerator and the denominator:
\[ = \sqrt{\frac{(s-c)(s-b)}{s^2}} \]

5. **Final Simplification**:
Thus, we can simplify further:
\[ = \frac{\sqrt{s-c}}{s} \]

6. **Expressing in Terms of Semi-Perimeter**:
Recall that \( s = \frac{a + b + c}{2} \) and \( s - c = \frac{a + b - c}{2} \):
\[ \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) = \frac{\sqrt{\frac{a + b - c}{2}}}{\frac{a + b + c}{2}} = \frac{a + b - c}{a + b + c} \]

### Final Answer:
\[ \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) = \frac{a + b - c}{a + b + c} \]