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 step-by-step, we start with the given conditions:

1. **Given that x, y, z are in Arithmetic Progression (AP)**:
- This means that \( 2y = x + z \).
- Let's call this Equation (1).

2. **Also given that \( \tan^{-1}x, \tan^{-1}y, \tan^{-1}z \) are in AP**:
- This means that \( 2\tan^{-1}y = \tan^{-1}x + \tan^{-1}z \).
- Let's call this Equation (2).

3. **Using the identity for \( 2\tan^{-1}y \)**:
- We know that \( 2\tan^{-1}y = \tan^{-1}\left(\frac{2y}{1 - y^2}\right) \).
- Therefore, from Equation (2), we can write:
\[ \tan^{-1}\left(\frac{2y}{1 - y^2}\right) = \tan^{-1}\left(\frac{x + z}{1 - xz}\right) \]

4. **Equating the arguments of the tangent inverse**:
- Since the tangent inverse function is one-to-one, we can equate the arguments:
\[ \frac{2y}{1 - y^2} = \frac{x + z}{1 - xz} \]

5. **Cross-multiplying to eliminate the fractions**:
- We get:
\[ 2y(1 - xz) = (x + z)(1 - y^2) \]

6. **Expanding both sides**:
- Left side: \( 2y - 2yxz \)
- Right side: \( x + z - xy^2 - zy^2 \)
- Thus, we have:
\[ 2y - 2yxz = x + z - xy^2 - zy^2 \]

7. **Rearranging the equation**:
- Bringing all terms to one side gives:
\[ 2y + xy^2 + zy^2 - 2yxz - x - z = 0 \]

8. **Substituting \( y = \frac{x + z}{2} \) from Equation (1)**:
- Substitute \( y \) into the equation:
\[ 2\left(\frac{x + z}{2}\right) + x\left(\frac{x + z}{2}\right)^2 + z\left(\frac{x + z}{2}\right)^2 - 2\left(\frac{x + z}{2}\right)xz - x - z = 0 \]

9. **Simplifying the equation**:
- This will lead to a quadratic equation in terms of \( x \) and \( z \). After simplification, we will find:
\[ (x - z)^2 = 0 \]

10. **Conclusion**:
- This implies \( x - z = 0 \) or \( x = z \).
- Since \( y = \frac{x + z}{2} \), we can conclude that \( x = y = z \).

Thus, the final answer is:
\[ x = y = z \]