JEE Main 2025 · 29 January · morning · Single correct · +4 / −1
Define a relation R on the interval [0, π/2) by x R y if and only if sec² x - tan² y = 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
Answer and solution
Answer (d) an equivalence relation
Solution
Maths sir · 4 steps
Swap sec squared for 1 plus tan squared, and this relation turns out to be plain equality.
- Rewrite the condition. Use sec² x = 1 + tan² x. Then sec² x - tan² y = 1 becomes 1 + tan² x - tan² y = 1, that is, tan² x = tan² y.
x R y ⇔ tan² x = tan² y ⇔ tan x = tan y ⇔ x = y
- Use the interval. On [0, π/2), tan x ≥ 0, so equal squares mean equal values. Also tan is strictly increasing there, so tan x = tan y forces x = y.
- Check the three properties. R is just x = y. Every x equals itself (reflexive). If x = y then y = x (symmetric). If x = y and y = z then x = z (transitive).
- So. R is reflexive, symmetric and transitive, so it is an equivalence relation: option (d).
The idea to keep. When a relation is defined by an equation, simplify it first. If it reduces to f(x) = f(y) for some function f, the relation is always an equivalence relation.
Answer from the NTA final answer key. Solution by PrepWiser.