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Math Help - questions about proving of identities for trigonometry

  1. #1
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    questions about proving of identities for trigonometry

    there are 2 questions.

    Prove the following identities:

    1.
    cosec^4 Ѳ – cot^4 Ѳ = cosec^2 Ѳ + cot^2 Ѳ


    2.
    2 tan A = [(cos A)/(cosec A – 1)] + [(cos A)/(cosec A +1)]




    for question 1, i start from the LHS, but get "stuck".
    for question 2, i start from the RHS, but also get "stuck".
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  2. #2
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    Quote Originally Posted by wintersoltice View Post
    there are 2 questions.

    Prove the following identities:

    1.
    cosec^4 Ѳ – cot^4 Ѳ = cosec^2 Ѳ + cot^2 Ѳ


    2.
    2 tan A = [(cos A)/(cosec A – 1)] + [(cos A)/(cosec A +1)]




    for question 1, i start from the LHS, but get "stuck".
    for question 2, i start from the RHS, but also get "stuck".
    Q1: LHS factorises as a difference of two squares. One of those factors is equal to 1 from to a standard identity.


    Q2: Replace cosec A with 1/sin A in RHS. Multiply top and bottom of each term by sin A to get

    sin A cos A/(1 - sin A) + sin A cos A/(1 + sin A).

    Get a common denominator - from a standard identity it reduces to cos^2 A. Combine the fractions etc.
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  3. #3
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    Hello, wintersoltice!

    Prove the following identities:

    [1] . csc^4(θ) – cot^4(θ) .= .cscē(θ) + cotē(θ)
    Factor the left side:

    csc^4(θ) - cot^4(θ) . = .[cscē(θ) - cotē(θ)] [cscē(θ) + cotē(θ)]
    . - . - . - . . . . . . . . . . . .\____________/
    . . . . . . . . . . . . . . . . . . . . .
    This is 1

    Therefore, we have: .cscē(θ) + cotē(θ)




    [2] . 2·tan A .= .[(cos A)/(csc A – 1)] + [(cos A)/(csc A +1)]
    On the right side, combine the fractions:

    . cos A - - csc A + 1 . . . . .cos A . . . csc A - 1
    ---------- . ----------- . + . ----------- . -----------
    csc A - 1 . csc A + 1 . . . .csc A + 1 - csc A - 1


    . . . . .cos A(csc A + 1) + cos A(csc A - 1)
    . . = . ------------------------------------------
    . . . . . . . . . .(csc A - 1)(csc A + 1)


    . . . . . 2·cos A·csc A . . . . 2·cot A . . . . . .2
    . . = . ----------------- . = . --------- . = . ------ . = . 2·tan A
    . . . . . . cscēA - 1 . . . . . . .cotēA . . . . .cot A

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  4. #4
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    Smile

    1.)
    making into a^2-b^2=(a+b)(a-b)

    left side part
    cosec^4thet+cot^4theta
    =(cosec^2theta+cot^2theta)(cosec^2-cot^2theta)
    =cosec^2theta+cot^2theta
    which is the right part

    2.)

    take right side part

    =(cosA/coesecA-1)+(cosA/cosecA+1)
    =cosA[cosecA-1+cosecA+1]/cosec^2-1
    =2cosAcosecA/cot^2A
    =2cotA/cot^A
    =2/CotA
    =2tanA which is the left part
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