If x,y,z are in AP and `tan^(-1)x,tan^(-1)y,tan^(-1)z` are also in AP, then

Options

A. `2x=3y=6z`
B. `6x=3y=2z`
C. `6x=4y=3z`
D. `x=y=z`

Explanation

To solve the problem, we need to show that if \(x, y, z\) are in arithmetic progression (AP) and \(\tan^{-1}x, \tan^{-1}y, \tan^{-1}z\) are also in AP, then \(x = y = z\).

### Step-by-Step Solution:

1. **Understanding AP**:
Since \(x, y, z\) are in AP, we can express them in terms of \(y\) and a common difference \(d\):
\[ x = y - d \quad \text{and} \quad z = y + d \]

2. **Using the property of AP for \(\tan^{-1}\)**:
Given that \(\tan^{-1}x, \tan^{-1}y, \tan^{-1}z\) are also in AP, we can use the property of AP:
\[ 2\tan^{-1}y = \tan^{-1}x + \tan^{-1}z \]

3. **Applying the formula for \(\tan^{-1}\)**:
We will use the formula for the tangent of a sum:
\[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \quad \text{(if } ab < 1\text{)} \]
Therefore, we have:
\[ \tan^{-1}x + \tan^{-1}z = \tan^{-1}\left(\frac{x + z}{1 - xz}\right) \]
Substituting \(x\) and \(z\):
\[ \tan^{-1}\left(\frac{(y - d) + (y + d)}{1 - (y - d)(y + d)}\right) = \tan^{-1}\left(\frac{2y}{1 - (y^2 - d^2)}\right) \]

4. **Setting up the equation**:
Now we equate the two expressions:
\[ 2\tan^{-1}y = \tan^{-1}\left(\frac{2y}{1 - y^2 + d^2}\right) \]
Taking tangent on both sides:
\[ \tan(2\tan^{-1}y) = \frac{2y}{1 - y^2 + d^2} \]
Using the double angle formula:
\[ \tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta} \]
We have:
\[ \tan(2\tan^{-1}y) = \frac{2y}{1 - y^2} \]
Thus, equating gives:
\[ \frac{2y}{1 - y^2} = \frac{2y}{1 - y^2 + d^2} \]

5. **Cross-multiplying**:
Cross-multiplying leads to:
\[ 2y(1 - y^2 + d^2) = 2y(1 - y^2) \]
If \(y \neq 0\), we can divide both sides by \(2y\):
\[ 1 - y^2 + d^2 = 1 - y^2 \]
This simplifies to:
\[ d^2 = 0 \]

6. **Conclusion**:
Since \(d^2 = 0\), it implies that \(d = 0\). Therefore, we conclude that:
\[ x = y = z \]

### Final Answer:
Thus, if \(x, y, z\) are in AP and \(\tan^{-1}x, \tan^{-1}y, \tan^{-1}z\) are also in AP, then \(x = y = z\).