`1-tan""A/2.tan""B/2=`

Options

A. `(2a)/(Sigma a)`
B. `(2b)/(Sigma a)`
C. `(2c)/(Sigma a)`
D. 0

Answer

To solve the problem \( 1 - \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) \), we will use the properties of triangles and some trigonometric identities.

### Step-by-Step Solution:

1. **Use the Half-Angle Tangent Formula:**
The half-angle formula for tangent states that:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} \]
where \( s \) is the semi-perimeter of the triangle, given by \( s = \frac{a+b+c}{2} \).

2. **Apply the Formula for Both Angles:**
Similarly, we can write:
\[ \tan\left(\frac{B}{2}\right) = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}} \]

3. **Substituting into the Expression:**
Now, substituting these into our expression:
\[ 1 - \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) = 1 - \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} \cdot \sqrt{\frac{s(s-b)}{(s-a)(s-c)}} \]

4. **Simplifying the Product:**
The product simplifies to:
\[ \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) = \frac{s(s-a)}{(s-b)(s-c)} \cdot \frac{s(s-b)}{(s-a)(s-c)} = \frac{s^2(s-a)(s-b)}{(s-a)(s-b)(s-c)(s-c)} \]
Thus, we have:
\[ 1 - \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) = 1 - \frac{s^2}{(s-c)(s)} = 1 - \frac{s}{s-c} \]

5. **Final Simplification:**
This simplifies to:
\[ = \frac{(s-c) - s}{s-c} = \frac{-c}{s-c} \]

6. **Expressing in Terms of Semi-Perimeter:**
Since \( s = \frac{a+b+c}{2} \), we can express \( s-c \) as:
\[ s - c = \frac{a+b-c}{2} \]
Therefore, the expression becomes:
\[ 1 - \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) = \frac{-c}{\frac{a+b-c}{2}} = \frac{-2c}{a+b-c} \]

### Final Result:
Thus, the final result is:
\[ 1 - \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) = \frac{-2c}{a+b-c} \]