sandesh sitaula

if sin2x=3sin2y then sin2x/sin2y=3/1 sin2x+sin2y/sin2x-sin2y=3+1/3-1 (using conponendo and dividend rule) now using formula of sin2x+sin2y and sin2x-sin2y 2sin (x+y)×cos (x-y)/2cos (x+y)×sin (x-y)=4/2 solving this we get the result