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math

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how do I simplify these?

1. (Cot/1-tan) + (Tan/1-Cot) - Tan - Cot


2. (1+cos) (csc-cot)

  • math -

    sloppy notation.

    the cot of what, the sin of what?
    sin, cos, tan, etc are mathematical operators
    I will use x as the "angle"

    (Cotx/1-tanx) + (Tanx/1-Cotx) - Tanx - Cotx

    I usually change everybody to sines and cosines, so ...

    = (cosx/sinx)/(1 - sinx/cosx) + (sinx/cosx)/(1 - cosx/sinx) - sinx/cox - cosx/sinx
    = (cos^2 x)/(sinx(cosx - sinx)) + (sin^2x)/(cosx(sinx-cosx) - sinx/cosx - cosx/sinx

    = (cos^2 x)/(sinx(cosx - sinx)) - (sin^2x)/(cosx(cosx - sinx) - sinx/cosx - cosx/sinx

    = .lots of messy typing here

    form a common denominator of
    (sinx)(cosx)(cosx-sinx) and try to finish it.

  • math -

    for the second, I would use the same approach.

    (1+cosx) (cscx-cotx)
    = (1 + cosx)(1/sinx - cosx/sinx)
    = 1/sinx - cosx/sinx + cosx/sinx - cos^2x/sinx
    = 1/sinx(1 - cos^2x)
    = 1/sinx(sin^2x_
    = sinx

  • math -

    (cos/sin/(1 - sin/cos) + sin/cos/(1-cos/sin) - sin/cos - cos /sin
    multiply top and bottom of all by sin cos
    cos^2/(sin cos -sin^2) + sin^2/(sin cos - cos^2) - (sin^2+cos^2)/sin cos

    cos^2/(sin cos -sin^2) + sin^2/(sin cos - cos^2) - 1 /sin cos

    cos^2/(sin (cos -sin)) - sin^2/(cos(cos - sin)) - 1 /sin cos

    writing cos as c and sin as s
    just doing first two terms for now
    c^2/(s(c-s)) - s^2/(c(c-s))

    1/(c-s) * (c^2/s -s^2/c)
    1/(c-s) * ( (c^3-s^3)/sc)
    1/(c-s) * ((c-s)(c^2 + sc + s^2)/sc
    (1+sc)/sc
    now put that -1/sc back
    (1+sc)/sc - 1/sc
    sc/sc
    1 Caramba !!!!

  • math -

    It is the sins, cos, etc of theta

  • math -

    OK, I was not about to type theta all the time either, and in fact even got tired of typing sin and cos.

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