Write #cos^4x-tan^2x# in terms of (a) #cos^2x# and (b) #sin^2x#?

Question

1213 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-14T14:23:03

Please see below.

In terms of #cos^2x#

#cos^4x-tan^2x#

= #(cos^2x)^2-sin^2x/cos^2x#

= #(cos^2x)^2-(1-cos^2x)/cos^2x#

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

In terms of #sin^2x#

#cos^4x-tan^2x#

= #(cos^2x)^2-sin^2x/cos^2x#

= #(1-sin^2x)^2-sin^2x/(1-sin^2x)#