Let $$P$$ be the relation defined on the set of all real numbers such that $$P = \{(a, b) : \sec^2 a - \tan^2 b = 1\}$$. Then, $$P$$ is:

The relation $$P$$ is defined on the set of all real numbers as $$P = \{(a, b) : \sec^2 a - \tan^2 b = 1\}$$. However, $$\sec a$$ and $$\tan b$$ are undefined when $$a$$ or $$b$$ is an odd multiple of $$\pi/2$$, i.e., $$a = (2k+1)\pi/2$$ or $$b = (2k+1)\pi/2$$ for any integer $$k$$. Therefore, the domain $$D$$ for which the relation is defined consists of all real numbers except these points. We will check the properties of reflexivity, symmetry, and transitivity for $$P$$ on this domain $$D$$.

Reflexivity: A relation is reflexive if for every $$a$$ in the domain, $$(a, a)$$ is in $$P$$. For any $$a \in D$$, we need to check if $$\sec^2 a - \tan^2 a = 1$$. Using the trigonometric identity $$\sec^2 \theta - \tan^2 \theta = 1$$, which holds for all $$\theta$$ where both functions are defined, we have $$\sec^2 a - \tan^2 a = 1$$. Thus, $$(a, a) \in P$$ for all $$a \in D$$. Therefore, $$P$$ is reflexive.

Symmetry: A relation is symmetric if whenever $$(a, b) \in P$$, then $$(b, a) \in P$$. Assume $$(a, b) \in P$$, so $$\sec^2 a - \tan^2 b = 1$$. We need to show that $$\sec^2 b - \tan^2 a = 1$$. From the given condition, $$\sec^2 a - \tan^2 b = 1$$. Rearranging, $$\sec^2 a = 1 + \tan^2 b$$. Using the identity $$\sec^2 b = 1 + \tan^2 b$$, we substitute to get $$\sec^2 b = 1 + \tan^2 b$$. Similarly, $$\tan^2 a = \sec^2 a - 1 = (1 + \tan^2 b) - 1 = \tan^2 b$$. Now, $$\sec^2 b - \tan^2 a = (1 + \tan^2 b) - \tan^2 a$$. Since $$\tan^2 a = \tan^2 b$$, this becomes $$1 + \tan^2 b - \tan^2 b = 1$$. Thus, $$\sec^2 b - \tan^2 a = 1$$, so $$(b, a) \in P$$. Therefore, $$P$$ is symmetric.

Transitivity: A relation is transitive if whenever $$(a, b) \in P$$ and $$(b, c) \in P$$, then $$(a, c) \in P$$. Assume $$(a, b) \in P$$ and $$(b, c) \in P$$, so $$\sec^2 a - \tan^2 b = 1$$ and $$\sec^2 b - \tan^2 c = 1$$. We need to show $$\sec^2 a - \tan^2 c = 1$$. From $$(a, b) \in P$$, $$\sec^2 a - \tan^2 b = 1$$, so $$\sec^2 a = 1 + \tan^2 b$$. From $$(b, c) \in P$$, $$\sec^2 b - \tan^2 c = 1$$, so $$\tan^2 c = \sec^2 b - 1$$. Now, $$\sec^2 a - \tan^2 c = (1 + \tan^2 b) - (\sec^2 b - 1)$$. Since $$\sec^2 b = 1 + \tan^2 b$$, substitute to get $$\sec^2 a - \tan^2 c = 1 + \tan^2 b - [(1 + \tan^2 b) - 1] = 1 + \tan^2 b - [\tan^2 b] = 1$$. Thus, $$\sec^2 a - \tan^2 c = 1$$, so $$(a, c) \in P$$. Therefore, $$P$$ is transitive.

Since $$P$$ is reflexive, symmetric, and transitive on its domain, it is an equivalence relation.

Hence, the correct answer is Option D.