We are given three positive real numbers, x, y, and z, with two conditions:

1. x, y, z are in Geometric Progression (GP).
2. \(\tan^{-1}x\), \(\tan^{-1}y\), \(\tan^{-1}z\) are in Arithmetic Progression (AP).

Let's analyze each condition.

Analyzing the Geometric Progression (GP) Condition for Positive Numbers

If x, y, and z are in GP, the ratio of consecutive terms is constant. For positive numbers, this means:

\begin{equation} \frac{y}{x} = \frac{z}{y} \implies y^2 = xz \quad (*)\end{equation}

Analyzing the Arithmetic Progression (AP) Condition for Inverse Tangents

If \(\tan^{-1}x\), \(\tan^{-1}y\), and \(\tan^{-1}z\) are in AP, the difference between consecutive terms is constant. For three terms, the middle term is the average of the other two:

\begin{equation} \tan^{-1}y - \tan^{-1}x = \tan^{-1}z - \tan^{-1}y \implies 2 \tan^{-1}y = \tan^{-1}x + \tan^{-1}z \quad (**)\end{equation}

Combining the Conditions using Tangent Identities

To combine the conditions, we can use properties of the \(\tan^{-1}\) function. Applying the tangent function to both sides of equation (**) gives:

\(\tan(2 \tan^{-1}y) = \tan(\tan^{-1}x + \tan^{-1}z)\)

Using the standard identities for \(\tan(2A) = \frac{2A}{1-A^2}\) and \(\tan(A+B) = \frac{A+B}{1-AB}\), we get:

\begin{equation} \frac{2y}{1-y^2} = \frac{x+z}{1-xz} \end{equation}

Note: These standard identities are applicable under certain conditions related to the arguments, but let's proceed with the algebraic result for now.

From the GP condition (*), we know \(y^2 = xz\). Substitute this into the equation:

\begin{equation} \frac{2y}{1-y^2} = \frac{x+z}{1-y^2} \end{equation analogous equation is derived even when \(y^2>1\) and \(xz>1\) using the appropriate \(\tan^{-1}\) identities involving \(\pi\)).

Assuming \(1-y^2 \neq 0\) (i.e., \(y \neq 1\)), we can multiply both sides by \((1-y^2)\):

\begin{equation} 2y = x+z \end{equation}

So, if \(y \neq 1\), the conditions imply that x, y, and z are in Arithmetic Progression (AP) as well as Geometric Progression (GP).

Positive Numbers in both AP and GP

A fundamental property in mathematics states that if three positive real numbers are in both AP and GP, they must be equal. Let the numbers be x, y, z. Since they are in GP, we have \(y^2=xz\). Since they are also in AP, we have \(2y=x+z\). Combining these equations:

From \(2y = x+z\), we have \(x = 2y - z\). Substitute this into \(y^2 = xz\):

\(y^2 = (2y-z)z\)

\(y^2 = 2yz - z^2\)

\(y^2 - 2yz + z^2 = 0\)

\((y-z)^2 = 0\)

This implies \(y=z\). Substituting \(y=z\) back into \(2y=x+z\), we get \(2y = x+y\), which implies \(y=x\). Therefore, \(x=y=z\).

This derivation holds under the assumption \(y \neq 1\).

Considering the case y = 1

If \(y=1\), the GP condition \(y^2=xz\) becomes \(1^2=xz\), so \(xz=1\). The AP condition \(2 \tan^{-1}y = \tan^{-1}x + \tan^{-1}z\) becomes \(2 \tan^{-1}1 = \tan^{-1}x + \tan^{-1}z\). \(2(\pi/4) = \tan^{-1}x + \tan^{-1}z\), which means \(\pi/2 = \tan^{-1}x + \tan^{-1}z\). For positive real numbers x and z, the identity \(\tan^{-1}x + \tan^{-1}z = \pi/2\) holds if and only if \(xz=1\). This is consistent with the GP condition when \(y=1\). So, the conditions are satisfied if \(y=1\) and \(xz=1\). While this case includes \(x=y=z=1\), it also includes possibilities where \(x \ne y \ne z\) (e.g., x=2, y=1, z=1/2). However, the fundamental relationship derived from the general case (\(y \ne 1\)) is \(x=y=z\). The structure of the problem and options suggests this is the required answer.

Conclusion

Based on the derivation from the GP and AP conditions for positive numbers, the relation \(x=y=z\) is obtained. While a specific case (\(y=1, xz=1\)) exists, the general algebraic outcome points towards equality.

Let's review the given options:

Options 3 and 4 contradict the GP condition for positive numbers unless they are all equal. Option 2 (\(xz=1\)) is true in the specific case \(y=1\), but not in all cases satisfying the conditions (e.g., x=y=z=2 gives \(xz=4\)). Option 1 (\(x=y=z\)) is the relationship derived from the general case where x, y, z are in both GP and AP.

Thus, the correct option is \(x=y=z\).

Revision Table: Key Conditions and Implications

Additional Information: Link Between Progressions

This problem highlights a classic result: positive numbers that are simultaneously in Geometric Progression and Arithmetic Progression must be equal. The \(\tan^{-1}\) condition, when translated using trigonometric identities, leads back to the AP condition (\(2y=x+z\)) for the numbers x, y, and z themselves, provided certain denominators are non-zero (\(1-y^2 \ne 0\)). The case where these denominators are zero corresponds to \(y=1\), which leads to \(xz=1\). However, the structure of the problem implies a singular correct answer among the options, suggesting the general result \(x=y=z\) is the intended conclusion.