If $$x$$, $$y$$, $$z$$ are positive numbers in A.P. and $$\tan^{-1}x$$, $$\tan^{-1}y$$ and $$\tan^{-1}z$$ are also in A.P., then which of the following is correct.

We are told that the three positive numbers $$x$$, $$y$$ and $$z$$ are in an arithmetic progression. By definition of an A.P. we therefore have

$$y=\dfrac{x+z}{2}.$$

Next, the three angles $$\tan^{-1}x$$, $$\tan^{-1}y$$ and $$\tan^{-1}z$$ are also in an arithmetic progression. For any three angles $$A,B,C$$ to be in A.P. we must have $$2B=A+C$$. Applying this to the present case gives

$$2\,\tan^{-1}y=\tan^{-1}x+\tan^{-1}z.$$

Now we take the tangent of both sides. Before doing so, let us recall the two standard trigonometric identities we shall need:

1. Sum formula for tangent: $$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\,\tan\beta}.$$

2. Double-angle formula for tangent: $$\tan(2\theta)=\dfrac{2\tan\theta}{1-\tan^{2}\theta}.$$

Using the double-angle formula on the left-hand side with $$\theta=\tan^{-1}y$$ gives

$$\tan\!\bigl(2\,\tan^{-1}y\bigr)=\dfrac{2y}{1-y^{2}}.$$

Using the sum formula on the right-hand side with $$\alpha=\tan^{-1}x$$ and $$\beta=\tan^{-1}z$$ yields

$$\tan\!\bigl(\tan^{-1}x+\tan^{-1}z\bigr)=\dfrac{x+z}{1-xz}.$$

Equating the two tangent values obtained from the equality of the angles, we have

$$\dfrac{2y}{1-y^{2}}=\dfrac{x+z}{1-xz}.$$

But from the A.P. condition we already know $$x+z=2y,$$ so we substitute this into the numerator on the right-hand side:

$$\dfrac{2y}{1-y^{2}}=\dfrac{2y}{1-xz}.$$

Both numerators are equal and non-zero (all numbers are positive), so we can cancel the common factor $$2y$$, obtaining

$$\dfrac{1}{1-y^{2}}=\dfrac{1}{1-xz}.$$

Cross-multiplying gives

$$1-xz=1-y^{2}\quad\Longrightarrow\quad y^{2}=xz.$$

Thus $$y$$ is the geometric mean of $$x$$ and $$z$$, while we already have $$y$$ as the arithmetic mean of $$x$$ and $$z$$. Let us now combine the two conditions.

From the arithmetic mean relation $$y=\dfrac{x+z}{2}$$, squaring both sides gives

$$y^{2}=\dfrac{(x+z)^{2}}{4}.$$

Substituting the geometric-mean equality $$y^{2}=xz$$ into this squared expression, we get

$$xz=\dfrac{(x+z)^{2}}{4}.$$

Multiplying throughout by $$4$$ yields

$$4xz=(x+z)^{2}=x^{2}+2xz+z^{2}.$$

Bringing all terms to one side, we have

$$0=x^{2}+2xz+z^{2}-4xz=x^{2}-2xz+z^{2}=(x-z)^{2}.$$

The square of a real number is zero only when the number itself is zero. Hence

$$x-z=0\quad\Longrightarrow\quad x=z.$$

Substituting $$x=z$$ back into $$y=\dfrac{x+z}{2}$$ gives

$$y=\dfrac{x+x}{2}=x.$$

Therefore $$x=y=z.$$

Looking at the options, we see that Option C states exactly this equality.

Hence, the correct answer is Option C.