Question

Question #0166b

Answer 1

`( 1-sin^4x+cos^4x) / (cos^2xsin^2x)`

`=( 1-(sin^4x-cos^4x)) / (cos^2xsin^2x)`

`=( 1-(sin^2x-cos^2x)(sin^2x+cos^2x))/ (cos^2xsin^2x)`

`=( 1-sin^2x+cos^2x) / (cos^2sin^2x)`

`=( cos^2x+cos^2x) / (cos^2sin^2x)`

`=( 2cos^2x) / (cos^2sin^2x)`

`=2 / sin^2x=2csc^2x`