Cosx-sinx/cosx+sinx simplify?

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Answer 1 · 2018-02-22T04:40:58

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

$"One way to simplify this could be:"$

${cosx-sinx}/{cosx+sinx}={cosx-sinx}/{cosx+sinx} cdot {cosx-sinx}/{cosx-sinx}$

$\qquad \qquad \qquad ={(cosx-sinx)^2}/{(cosx-sinx)(cosx+sinx)}$

$\qquad \qquad \qquad ={cos^2x-2cosxsinx+sin^2x}/{cos^2x-sin^2x}$

$"[now recall that:" \ 2cosxsinx=sin2x; cos^2x-sin^2x=cos2x"]"$

$\qquad \qquad \qquad ={(sin^2x+cos^2x)-sin2x}/{cos2x}$

$"[now recall that:" \ quad sin^2x+cos^2x=1"]"$

$\qquad \qquad \qquad ={1-sin2x}/{cos2x}$

$\qquad \qquad \qquad ={1}/{cos2x} - {sin2x}/{cos2x}$

$\qquad \qquad \qquad =sec2x-tan2x.$

$"Thus, we have:"$

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