#"One way to simplify this could be:"#
#{cosx-sinx}/{cosx+sinx}={cosx-sinx}/{cosx+sinx} cdot {cosx-sinx}/{cosx-sinx}#
# \qquad \qquad \qquad ={(cosx-sinx)^2}/{(cosx-sinx)(cosx+sinx)} #
# \qquad \qquad \qquad ={cos^2x-2cosxsinx+sin^2x}/{cos^2x-sin^2x} #
#"[now recall that:" \ 2cosxsinx=sin2x; cos^2x-sin^2x=cos2x"]" #
# \qquad \qquad \qquad ={(sin^2x+cos^2x)-sin2x}/{cos2x} #
#"[now recall that:" \ quad sin^2x+cos^2x=1"]" #
# \qquad \qquad \qquad ={1-sin2x}/{cos2x} #
# \qquad \qquad \qquad ={1}/{cos2x} - {sin2x}/{cos2x}#
# \qquad \qquad \qquad =sec2x-tan2x. #
#"Thus, we have:"#
# \qquad \qquad \qquad \qquad \qquad \qquad {cosx-sinx}/{cosx+sinx}=sec2x-tan2x.#