Prove that:sec^2x+cosec^2x>=4

sin^2A cosec^2A=>2​
  • 0
L.H.S: sec^2x+ cosec^2x =(1+tan^2x)+(1+cot^2x) =2+tan^2x+cot^2x = 2+ sin^2x / cos^2x +cos^2x/sin^2x = 2+(sin^4x+cos^4x)/sin^2x cos^2x =2+ [(cos^2x-sin^2x)2+2sin^2x cos^2x]/sin^2x cis^2x =2+(cos^2 2x)/sin^2x cos^2x +2 = 4+cos^2 2x/sin^2x cos^2x Because max.value of sinx and cosx is 1. Therefore, 4+cos^2 2x/sin^2x cos^2x>4

  • 5
i dont know
 
  • -4
Sin^2A Cosec^2A=2
  • -2
What are you looking for?