Cosx-sinx/cosx+sinx simplify?

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Answer 1 · 2018-02-22T04:40:58

# \qquad \qquad \qquad \qquad {cosx-sinx}/{cosx+sinx}=sec2x-tan2x.#

#"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.#