# trigonometry

1/sin2A+cos4A/sin4A

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1. 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

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2. 1/sin2x+2(cos2x)^2 -1/2sin2xcos2x
= 1/sin2x[2cos2x+2(cos2x)^2 - 1]/2cos2x
=[2cos2x+2(cos2x)^2 -1]/sin4x

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