Key Concepts and Formulas

• Trigonometric Identity: $\sec^2 \theta - \tan^2 \theta = 1$. This identity holds for all $\theta$ where $\sec \theta$ and $\tan \theta$ are defined.
• Properties of Relations:
  • Reflexive: A relation $R$ on a set $A$ is reflexive if $x R x$ for all $x \in A$.
  • Symmetric: A relation $R$ on a set $A$ is symmetric if for all $x, y \in A$, if $x R y$, then $y R x$.
  • Transitive: A relation $R$ on a set $A$ is transitive if for all $x, y, z \in A$, if $x R y$ and $y R z$, then $x R z$.
• Equivalence Relation: A relation that is reflexive, symmetric, and transitive.
• Properties of Tangent Function on $\left[0, \frac{\pi}{2}\right)$: The function $\tan \theta$ is non-negative and strictly increasing (one-to-one) on the interval $\left[0, \frac{\pi}{2}\right)$. This implies that if $\tan x = \tan y$ for $x, y \in \left[0, \frac{\pi}{2}\right)$, then $x = y$.

Step-by-Step Solution

Step 1: Simplify the defining condition of the relation $R$. The relation $R$ is defined on the interval $\left[0, \frac{\pi}{2}\right)$ by $x R y$ if and only if $\sec^2x - \tan^2y = 1$. We use the fundamental trigonometric identity $\sec^2 \theta = 1 + \tan^2 \theta$. Substituting this into the condition for $x$: $(1 + \tan^2x) - \tan^2y = 1$ Subtracting 1 from both sides, we get: $\tan^2x - \tan^2y = 0$ $\tan^2x = \tan^2y$ For $x, y \in \left[0, \frac{\pi}{2}\right)$, the tangent function is non-negative. Therefore, $\tan^2x = \tan^2y$ implies $\tan x = \tan y$. Since the tangent function is strictly increasing (one-to-one) on the interval $\left[0, \frac{\pi}{2}\right)$, the condition $\tan x = \tan y$ implies $x = y$. Thus, the relation $x R y$ is equivalent to the condition $x = y$ for all $x, y \in \left[0, \frac{\pi}{2}\right)$.

Step 2: Check for Reflexivity. A relation $R$ is reflexive if for every element $x$ in the set, $x R x$ holds. Using our simplified condition, $x R x$ means $x = x$. This statement is always true for any $x \in \left[0, \frac{\pi}{2}\right)$. Therefore, the relation $R$ is reflexive.

Step 3: Check for Symmetry. A relation $R$ is symmetric if for every pair of elements $x, y$ in the set, whenever $x R y$ holds, then $y R x$ also holds. Assume $x R y$. From our simplified condition, this means $x = y$. Now we need to check if $y R x$ holds. For $y R x$ to hold, we need $y = x$. Since the condition $x = y$ implies $y = x$ (equality is symmetric), the relation $R$ is symmetric.

Step 4: Check for Transitivity. A relation $R$ is transitive if for every triplet of elements $x, y, z$ in the set, whenever $x R y$ and $y R z$ hold, then $x R z$ also holds. Assume $x R y$ and $y R z$. From our simplified condition:

• $x R y$ implies $x = y$.
• $y R z$ implies $y = z$. If $x = y$ and $y = z$, then it logically follows that $x = z$. Now we need to check if $x R z$ holds, which means we need to check if $x = z$. Since $x = z$ is derived from the premises $x = y$ and $y = z$, the relation $R$ is transitive.

Step 5: Conclude the nature of the relation. Since the relation $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Common Mistakes & Tips

• Domain Sensitivity: Always pay close attention to the domain of the variables. In this problem, the interval $\left[0, \frac{\pi}{2}\right)$ is crucial for simplifying $\tan^2x = \tan^2y$ to $x=y$. If the domain were different, the relation might not simplify to equality.
• Trigonometric Identity Misapplication: Ensure the trigonometric identity $\sec^2 \theta - \tan^2 \theta = 1$ is applied correctly and that the functions $\sec \theta$ and $\tan \theta$ are defined within the given interval. For $\left[0, \frac{\pi}{2}\right)$, both are defined.
• Algebraic Simplification: Double-check all algebraic manipulations, especially when dealing with squares and square roots. In this case, $\tan^2x = \tan^2y \implies \tan x = \pm \tan y$. However, within the specified domain, $\tan x$ and $\tan y$ are non-negative, so $\tan x = \tan y$ is the only possibility.

Summary

The relation $R$ defined by $\sec^2x - \tan^2y = 1$ on the interval $\left[0, \frac{\pi}{2}\right)$ was first simplified. Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$, the condition $\sec^2x - \tan^2y = 1$ reduces to $\tan^2x = \tan^2y$. Given the domain $\left[0, \frac{\pi}{2}\right)$, where the tangent function is non-negative and one-to-one, this further simplifies to $x = y$. We then checked the properties of reflexivity, symmetry, and transitivity for the relation $x=y$. The relation $x=y$ is reflexive (as $x=x$), symmetric (as $x=y \implies y=x$), and transitive (as $x=y$ and $y=z \implies x=z$). Therefore, $R$ is an equivalence relation.

The final answer is $\boxed{\text{an equivalence relation}}$.