If `sin(y+z-x),sin(z+x-y),"sin"(x+y-z)` are in A.P., then `t a n x ,tany ,tanz` are in
(a)A.P.
(b) G.P.
(c) H.P.
(d) none of these

Answer

To solve the problem, we need to determine the relationship between \( \tan x \), \( \tan y \), and \( \tan z \) given that \( \sin(y + z - x) \), \( \sin(z + x - y) \), and \( \sin(x + y - z) \) are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the Condition of A.P.**:
Since \( \sin(y + z - x) \), \( \sin(z + x - y) \), and \( \sin(x + y - z) \) are in A.P., we can express this condition mathematically:
\[ 2 \sin(z + x - y) = \sin(y + z - x) + \sin(x + y - z) \]

2. **Using the Sine Difference Formula**:
We can apply the sine difference formula:
\[ \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]
Let \( A = y + z - x \) and \( B = x + y - z \). Then we can rewrite our equation:
\[ 2 \sin(z + x - y) = 2 \cos\left(\frac{(y + z - x) + (x + y - z)}{2}\right) \sin\left(\frac{(y + z - x) - (x + y - z)}{2}\right) \]

3. **Simplifying the Angles**:
Calculate \( A + B \) and \( A - B \):
- \( A + B = (y + z - x) + (x + y - z) = 2y \)
- \( A - B = (y + z - x) - (x + y - z) = 2z - 2x \)

Substituting these into the equation gives:
\[ 2 \sin(z + x - y) = 2 \cos(y) \sin(z - x) \]

4. **Dividing by 2**:
Simplifying further, we have:
\[ \sin(z + x - y) = \cos(y) \sin(z - x) \]

5. **Using the Sine and Cosine Relationships**:
Now, we can express the sine and cosine in terms of tangent:
\[ \tan(z + x - y) = \frac{\sin(z + x - y)}{\cos(z + x - y)} \]
and similarly for the other terms.

6. **Finding the Relationship**:
From the above equations, we can derive:
\[ \tan x + \tan z = 2 \tan y \]
This indicates that \( \tan x \), \( \tan y \), and \( \tan z \) are in A.P.

### Conclusion:
Thus, we conclude that \( \tan x \), \( \tan y \), and \( \tan z \) are in arithmetic progression (A.P.).

### Answer:
(a) A.P.