trigonometry

Prove:
tan2x+sec2x = cosx+sinx/ cosx-sinx

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  1. I am sure you meant :
    tan2x+sec2x = (cosx+sinx)/(cosx-sinx)

    LS
    = sin2x/cos2x + 1/cos2x
    = (sin2x +1)/cos2x
    = (2sinxcosx + sin^2x + cos^2x)/(cos^2x - sin^2x)
    = (sinx + cosx)^2/[(cosx-sinx)(cosx+sinx)]
    = (sinx + cosx)/(cosx - sinx)
    = RS

    Q.E.D.

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  2. This time start from the right-hand side by taking advantage of the term cos(x)-sin(x):
    (cos(x)+sin(x))/(cos(x)-sin(x))
    multiply top and bottom by cos(x)+sin(x)
    (cos(x)+sin(x))^sup2;/(cos²(x)-sin²(x)
    =(cos²(x)+sin²(x)+2sin(x)cos(x))/(cos²(x)-sin²(x))
    =(1+sin(2x))/cos(2x)
    =sec(2x)+tan(2x)

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  3. Thank you so much Reiny and Mathmate. Both of you guys gave me great ways to solving this problem.

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