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Math Help - Trigonometry Help

  1. #1
    Bar0n janvdl's Avatar
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    Trigonometry Help

    Here's a problem not even my teacher could figure out.

    Everyone knows that cot(90 + y) = -tan y

    So now we have to prove it.

    cot(90 + y) = 1/(tan 90 + y)

    LHS = {1 - (tan 90)(tan y)} / {(tan 90)(tan y}

    But what the heck are we to do with tan 90???
    And how will one prove this identity?
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  2. #2
    Member Glaysher's Avatar
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    Quote Originally Posted by janvdl View Post
    Here's a problem not even my teacher could figure out.

    Everyone knows that cot(90 + y) = -tan y

    So now we have to prove it.

    cot(90 + y) = 1/(tan 90 + y)

    LHS = {1 - (tan 90)(tan y)} / {(tan 90)(tan y}

    But what the heck are we to do with tan 90???
    And how will one prove this identity?
    cot (90 + y) = cos (90 + y)/sin (90 + y)

    = [cos 90 cos y - sin 90 sin y]/[sin 90 cos y + cos 90 sin y]

    = -sin y/cos y

    = -tan y
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  3. #3
    Super Member

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    Hello, janvdl!

    Prove: .cot(90 + θ) .= .- tanθ
    . . . . . . . . . . . . . . . . . . . . . . . . . . . .tanα + tanβ
    You know this formula: .tan(α + β) .= .----------------
    . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - tanαĚtanβ

    . . . . . . . . . . . . . . . . . . . . . . . . . . . . cotαĚcotβ - 1
    We can derive this one: .cot(α + β) .= .----------------
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . cotα + cotβ


    . . . . . . . . . . . . . . . . cot90Ěcotθ - 1 . . . 0Ěcotθ - 1 . . . .-1
    Then: .cot(90 + θ) .= .----------------- .= .------------ .= .------ .= .- tanθ
    . . . . . . . . . . . . . . . . .cot90 + cotθ . . . .0 + cotθ . . . .cotθ

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