Prove this Trig identity?

So I am trying to figure out this question, and I always get to a point where i get stuck, and dont know what to do... this is the question:

cos x-sin x/cos x+sin x = sec2x-tan2x

i tried working with the left side, but that didnt go so well for me.... what i had was:

cos^2x +sinx-sin^2x-cosx/cos^2x+sinx+sin^2x+cosx

2 Answers

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  • Mike G
    Lv 7
    4 years ago
    Favorite Answer

    I did it your way, Jeremiah but multiplied top and bottom by (cosx-sinx)

    (cosx - sinx) /(cosx + sinx)

    = (cosx - sinx)(cos-sinx) /[(cosx + sinx)(cosx-sinx)]

    = [cos^2(x)+sin^2(x)-2sinx*cosx]/[cos^2(x)-sin^2(x)]

    = [1-2sinx*cosx]]/cos(2x)

    = [1-sin(2x)]/cos(2x)

    = sec(2x) - tan(2x)

  • 4 years ago

    Hello,

    (cosx - sinx) /(cosx + sinx) = sec(2x) - tan(2x)

    let's write sec(2x) as 1 /cos(2x) and tan(2x) as sin(2x) /cos(2x):

    (cosx - sinx) /(cosx + sinx) = [1 /cos(2x)] - [sin(2x) /cos(2x)]

    (cosx - sinx) /(cosx + sinx) = [1 - sin(2x)] /cos(2x)

    let's apply the double-angle identity cos(2x) = cos²x - sin²x:

    (cosx - sinx) /(cosx + sinx) = [1 - sin(2x)] /(cos²x - sin²x)

    let's factor the denominator at the right side:

    (cosx - sinx) /(cosx + sinx) = [1 - sin(2x)] /[(cosx + sinx)(cosx - sinx)]

    let's apply the fundamental identity 1 = cos²x + sin²x:

    (cosx - sinx) /(cosx + sinx) = [(cos²x + sin²x) - sin(2x)] /[(cosx + sinx)(cosx - sinx)]

    (cosx - sinx) /(cosx + sinx) = [cos²x + sin²x - sin(2x)] /[(cosx + sinx)(cosx - sinx)]

    let's apply the double-angle identity sin(2x) = 2sinx cosx:

    (cosx - sinx) /(cosx + sinx) = (cos²x + sin²x - 2sinx cosx) /[(cosx + sinx)(cosx - sinx)]

    let's note that the numerator at the right side is a square:

    (cosx - sinx) /(cosx + sinx) = (cos²x - 2sinx cosx + sin²x) /[(cosx + sinx)(cosx - sinx)]

    (cosx - sinx) /(cosx + sinx) = (cosx - sinx)² /[(cosx + sinx)(cosx - sinx)]

    finally let's simplify:

    (cosx - sinx) /(cosx + sinx) = (cosx - sinx) /(cosx + sinx) (Q.E.D.)

    I hope it helps

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