In any triangle ABC ,`(tan(A/2)-tan(B/2))/(tan(A/2)+tan(B/2))` is equal to

Options

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

Answer

To solve the problem, we need to evaluate the expression \((\tan(A/2) - \tan(B/2)) / (\tan(A/2) + \tan(B/2))\) in the context of triangle \(ABC\).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression:
\[ \frac{\tan(A/2) - \tan(B/2)}{\tan(A/2) + \tan(B/2)} \]

2. **Using the Tangent Half-Angle Formula**:
Recall that:
\[ \tan(x) = \frac{\sin(x)}{\cos(x)} \]
Therefore, we can express \(\tan(A/2)\) and \(\tan(B/2)\) in terms of sine and cosine:
\[ \tan(A/2) = \frac{\sin(A/2)}{\cos(A/2)}, \quad \tan(B/2) = \frac{\sin(B/2)}{\cos(B/2)} \]

3. **Substituting into the Expression**:
Substitute these into the expression:
\[ \frac{\frac{\sin(A/2)}{\cos(A/2)} - \frac{\sin(B/2)}{\cos(B/2)}}{\frac{\sin(A/2)}{\cos(A/2)} + \frac{\sin(B/2)}{\cos(B/2)}} \]

4. **Finding a Common Denominator**:
The common denominator for both the numerator and denominator is \(\cos(A/2) \cos(B/2)\):
\[ = \frac{\sin(A/2) \cos(B/2) - \sin(B/2) \cos(A/2)}{\sin(A/2) \cos(B/2) + \sin(B/2) \cos(A/2)} \]

5. **Using the Sine Difference and Sum Formulas**:
The numerator can be simplified using the sine difference formula:
\[ \sin(A/2) \cos(B/2) - \sin(B/2) \cos(A/2) = \sin\left(\frac{A}{2} - \frac{B}{2}\right) \]
The denominator can be simplified using the sine sum formula:
\[ \sin(A/2) \cos(B/2) + \sin(B/2) \cos(A/2) = \sin\left(\frac{A}{2} + \frac{B}{2}\right) \]

6. **Final Expression**:
Thus, we can rewrite our expression as:
\[ \frac{\sin\left(\frac{A}{2} - \frac{B}{2}\right)}{\sin\left(\frac{A}{2} + \frac{B}{2}\right)} \]

7. **Using the Angle Sum Property**:
Since \(A + B + C = 180^\circ\) in triangle \(ABC\), we have:
\[ \frac{A + B}{2} = 90^\circ - \frac{C}{2} \]
Therefore,
\[ \sin\left(\frac{A}{2} + \frac{B}{2}\right) = \cos\left(\frac{C}{2}\right) \]

8. **Using the Sine Rule**:
By applying the sine rule, we know:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Thus, we can express \(\sin A\) and \(\sin B\) in terms of \(a\), \(b\), and \(c\):
\[ \sin A = \frac{a}{k}, \quad \sin B = \frac{b}{k}, \quad \sin C = \frac{c}{k} \]

9. **Final Result**:
After substituting these values back, we find that:
\[ \frac{\sin\left(\frac{A}{2} - \frac{B}{2}\right)}{\cos\left(\frac{C}{2}\right)} = \frac{a - b}{c} \]
Thus, the final answer is:
\[ \frac{a - b}{c} \]

### Conclusion:
The value of \(\frac{\tan(A/2) - \tan(B/2)}{\tan(A/2) + \tan(B/2)}\) is equal to \(\frac{a - b}{c}\).