If x, y, z are in G.P. and `tan^(-1) x, tan^(-1)y and tan^(-1)z` are in A.P., then

Options

A. x = y = z or y `ne` 1
B. `z = (1)/(x)`
C. x = y = z, but their common value is not necessarily zero
D. x = y = z = 0

Answer

To solve the problem, we need to establish the relationship between \(x\), \(y\), and \(z\) given that they are in geometric progression (G.P.) and that \(\tan^{-1} x\), \(\tan^{-1} y\), and \(\tan^{-1} z\) are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding G.P. Condition**:
Since \(x\), \(y\), and \(z\) are in G.P., we can express this condition mathematically:
\[ y^2 = xz \]

2. **Understanding A.P. Condition**:
Since \(\tan^{-1} x\), \(\tan^{-1} y\), and \(\tan^{-1} z\) are in A.P., we can express this condition as:
\[ 2 \tan^{-1} y = \tan^{-1} x + \tan^{-1} z \]

3. **Using the Formula for Sum of Inverses**:
We can use the formula for the sum of inverse tangents:
\[ \tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A + B}{1 - AB} \right) \quad \text{(provided } AB < 1\text{)} \]
Applying this to our A.P. condition:
\[ 2 \tan^{-1} y = \tan^{-1} \left( \frac{x + z}{1 - xz} \right) \]

4. **Expressing \(2 \tan^{-1} y\)**:
We can also express \(2 \tan^{-1} y\) using the double angle formula for tangent:
\[ 2 \tan^{-1} y = \tan^{-1} \left( \frac{2y}{1 - y^2} \right) \]

5. **Setting the Two Expressions Equal**:
Now we equate the two expressions:
\[ \tan^{-1} \left( \frac{2y}{1 - y^2} \right) = \tan^{-1} \left( \frac{x + z}{1 - xz} \right) \]

6. **Removing the Inverse Tangent**:
Since the tangent function is one-to-one in the relevant range, we can set the arguments equal to each other:
\[ \frac{2y}{1 - y^2} = \frac{x + z}{1 - xz} \]

7. **Substituting \(y^2 = xz\)**:
From the G.P. condition, we have \(y^2 = xz\). We can substitute this into our equation:
\[ \frac{2y}{1 - xz} = \frac{x + z}{1 - xz} \]

8. **Simplifying the Equation**:
Since \(1 - xz\) is common on both sides (and not equal to zero), we can simplify:
\[ 2y = x + z \]

9. **Conclusion**:
Now, we have established that \(2y = x + z\). This implies that \(x\), \(y\), and \(z\) are in arithmetic progression (A.P.) as well. Since \(x\), \(y\), and \(z\) are both in G.P. and A.P., it follows that they must be equal:
\[ x = y = z \]

### Final Answer:
Thus, we conclude that \(x = y = z\), where the common value is not necessarily zero.