Define a relation R on the interval $\left[0, \frac{\pi}{2}\right)$ by x R y if and only if $\sec ^{2} x-\tan ^{2} y=1$. Then $R$ is :

(B) both reflexive and transitive but not symmetric

(C) both reflexive and symmetric but not transitive

(D) reflexive but neither symmetric not transitive

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Sets Relations and Functions

Define a relation R on the interval

[0, \frac{π}{2})

by x R y if and only if

sec^{2} x - tan^{2} y = 1

. Then

R

is :

Answer: (A)

The defining condition is

sec^{2} x - tan^{2} y = 1

. Since

sec^{2} x = 1 + tan^{2} x

identically, this is the same as

1 + tan^{2} x - tan^{2} y = 1 ⟹ tan^{2} x = tan^{2} y .

[0, π /2)

tan

is non-negative and strictly increasing, so

tan^{2} x = tan^{2} y

forces

tan x = tan y

, hence

x = y

.

So

R

is exactly the relation

x = y

restricted to this interval: for any

x

x R x

holds (reflexive);

x R y \Rightarrow x = y \Rightarrow y R x

(symmetric); and

x R y, y R z \Rightarrow x = y = z \Rightarrow x R z

(transitive).

Hence

R

is an equivalence relation.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option