Question #200ab

Question

1061 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-02T21:25:32

see below

Left Side#= cos^4x (1+sin^4x/cos^4x)#

#=cos^4x xx (cos^4x+sin^4x)/cos^4x#

#=cos^4x+sin^4x#

#= cos^2x cos^2x+ sin^2x sin^2x#

#=(1-sin^2x)cos^2x+(1-cos^2x)sin^2x#

#=cos^2x-sin^2xcos^2x+sin^2x-sin^2xcos^2x#

#=cos^2x+sin^2x-2sin^2xcos^2x#

#=1-2sin^2xcos^2x#

#=#Right Side