Define a relation R on the interval $[0, π/2)$ by $x$ R $y$ if and only if $\sec^2x - \tan^2y = 1$. Then R is :
(A)
both reflexive and symmetric but not transitive
(B)
both reflexive and transitive but not symmetric
(C)
reflexive but neither symmetric not transitive
(D)
an equivalence relation
MEDIUM
Correct Answer:D
Explanation
To check if the relation R is reflexive, symmetric, and transitive:
1. **Reflexive:**
Replace $y$ with $x$ in the given relation: $\sec^2x - \tan^2x = 1$. This is a trigonometric identity, so the relation is reflexive.
2. **Symmetric:**
Given $\sec^2x - \tan^2y = 1$, we can rewrite this as $1 + \tan^2x - \tan^2y = 1$, which implies $\tan^2x = \tan^2y$. Also, $\sec^2x - \tan^2y = 1$ can be rewritten as $1 + \tan^2x - \sec^2y + 1 = 1$, leading to $\sec^2y - \tan^2x = 1$. Thus, if $xRy$, then $yRx$, so the relation is symmetric.
3. **Transitive:**
Suppose $\sec^2x - \tan^2y = 1$ and $\sec^2y - \tan^2z = 1$. Adding these two equations, we get:
$\sec^2x - \tan^2y + \sec^2y - \tan^2z = 1 + 1$
$\sec^2x + (\sec^2y - \tan^2y) - \tan^2z = 2$
Since $\sec^2y - \tan^2y = 1$, we have:
$\sec^2x + 1 - \tan^2z = 2$
$\sec^2x - \tan^2z = 1$
Thus, if $xRy$ and $yRz$, then $xRz$, so the relation is transitive.
Since the relation is reflexive, symmetric, and transitive, it is an equivalence relation.