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Trigonometry 19 Online
OpenStudy (anonymous):

cos2x/cosx-sinx -sinx

OpenStudy (anonymous):

that would be (cos^2 x-sin^2 x/cosx-sinx) -sinx => [(cosx+sinx) (cosx-sinx)/ cosx-sinx] -sinx => cosx+sinx - sinx = cosx.

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