In `DeltaABC`, If `x= tan((B-C)/2) tan(A/2),y= tan((C-A)/2)tan(B/2), z=tan((A-B)/2) tan(C/2)` , then `x+y+z` (in terms of x, y, z only) is

Answer

To solve the problem, we start with the definitions of \( x \), \( y \), and \( z \) in terms of the angles of triangle \( ABC \):

1. **Given Definitions**:
\[ x = \tan\left(\frac{B-C}{2}\right) \tan\left(\frac{A}{2}\right) \]
\[ y = \tan\left(\frac{C-A}{2}\right) \tan\left(\frac{B}{2}\right) \]
\[ z = \tan\left(\frac{A-B}{2}\right) \tan\left(\frac{C}{2}\right) \]

2. **Finding \( x + y + z \)**:
We need to express \( x + y + z \) in terms of \( x \), \( y \), and \( z \) only.

3. **Using the Tangent Half-Angle Identities**:
We can use the tangent subtraction formula:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
This will help us relate the terms.

4. **Expressing Each Term**:
Each term can be rewritten using the tangent subtraction identities:
\[ x = \frac{\tan A - \tan C}{1 + \tan A \tan C} \cdot \tan\left(\frac{A}{2}\right) \]
\[ y = \frac{\tan C - \tan A}{1 + \tan C \tan A} \cdot \tan\left(\frac{B}{2}\right) \]
\[ z = \frac{\tan A - \tan B}{1 + \tan A \tan B} \cdot \tan\left(\frac{C}{2}\right) \]

5. **Combining the Terms**:
Now, we can combine these expressions:
\[ x + y + z = \frac{\tan A - \tan C}{1 + \tan A \tan C} \tan\left(\frac{A}{2}\right) + \frac{\tan C - \tan A}{1 + \tan C \tan A} \tan\left(\frac{B}{2}\right) + \frac{\tan A - \tan B}{1 + \tan A \tan B} \tan\left(\frac{C}{2}\right) \]

6. **Finding a Common Denominator**:
The common denominator for these fractions is \( (1 + \tan A \tan C)(1 + \tan C \tan A)(1 + \tan A \tan B) \).

7. **Simplifying the Expression**:
After substituting and simplifying, we find that:
\[ x + y + z = \frac{-C^2 - A^2 - B^2 - AB - BC - CA}{ABC} \]
This can be expressed in terms of \( x, y, z \).

8. **Final Expression**:
Ultimately, we find that:
\[ x + y + z = \frac{xyz}{ABC} \]