WBJEE 2014 Maths Solved Paper

For reflexive, θ = φ so sec2θ – tan2θ = 1, Hence Reflexive
For symmetric, sec2θ – tan2φ = 1 so, (1 + tan2θ) – (sec2φ – 1) = 1 so, sec2φ – tan2θ = 1. Hence symmetric
For Transitive, let sec2θ – tan2φ = 1 and sec2φ – tan2γ = 1 so, 1 + tan2φ – tan2γ = 1 or, sec2θ – tan2γ = 1. Hence Transitive