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Calculus

Prove the identity
cscx+cotx-1/cotx-cscx+1 = 1+cosx/sinx

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Anonymous

  1. I shall assume you mean
    (cscx+cotx-1)/(cotx-cscx+1) = (1+cosx)/sinx
    If so, then multiplying on the left by (cotx+(cscx-1))/(cotx+(cscx-1) you get
    (cotx+(cscx-1))^2 / (cot^2x - (cscx-1)^2)
    = (cot^2x + 2cotx(cscx-1) + csc^2x - 2cscx + 1)/(cot^2x - csc^2x + 2cscx - 1)
    = (csc^2x-1 + 2cotx*cscx - 2cotx + csc^2x - 2cscx + 1)/(csc^2x-1 - csc^2x + 2cscx - 1)
    = (2csc^2x +2(cotxcscx-cotx-cscx))/(2cscx-2)
    = (csc^2x+cscx*cotx-cscx-cotx)/(cscx-1)
    = (cscx(cscx+cotx)-(cscx+cotx))/(cscx-1)
    = (cscx-1)(cscx+cotx)/(cscx-1)
    = cscx+cotx
    = (1+cosx)/sinx

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    Steve

  2. The way you typed it, it is not an identity.
    Nobody is going to start guessing at the many permutations that are
    possible in placing brackets.
    e.g. one could be
    cscx + (cotx-1)/(cotx-cscx+1) = (1+cosx)/sinx , but that would not be an
    identity.
    So... repost your equation, placing brackets in the correct places.

    Btw, before trying to prove that an identity is true, substitute some non-standard angle to see if it is true. I usually try 19.5° as a personal choice.
    If the equation shows up as true, more than likely, your identity is true.

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    Reiny

  4. ahh, I see that Steve correctly guessed what you meant.

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    Reiny

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