If you mean
1/sin2x + cos4x/sin4x
then that is
1/sin2x + (2cos^2 2x - 1)/(2sin2x cos2x)
= (2cos2x + 2cos^2 2x - 1)/(2sin2x cos 2x)
Not sure just what you are after
1/sin2A+cos4A/sin4A
2 answers
1/sin2x+2(cos2x)^2 -1/2sin2xcos2x
= 1/sin2x[2cos2x+2(cos2x)^2 - 1]/2cos2x
=[2cos2x+2(cos2x)^2 -1]/sin4x
= 1/sin2x[2cos2x+2(cos2x)^2 - 1]/2cos2x
=[2cos2x+2(cos2x)^2 -1]/sin4x