WBJEE 2013 Math Solved Paper

sin2a + cos2b = 1
Reflexive : sin2a + cos2a = 1
⇒ aRa
sin2a + cos2b = 1, 1 – cos2a + 1 – sin2b = 1
sin2b + cos2a = 1
⇒ bRa
Hence symmetric Let aRb bR
sin2a + cos2b = 1 ............. (1)
sin2b + cos2c = 1 ............... (2)
(1) + (2)
sin2a + cos2c = 1
Hence transitive therefore equivalence relation.