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Math Help - Prove that each of the following is an identity

  1. #1
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    Question Prove that each of the following is an identity

    I am frustated with these two problems, i can not figure them out, ive tried many identities but still no correct answer.

    The problems are
    #1.) (1+cot(x))^2-cot(x)=1/(1-cos(x))(1+cos(x))
    #2.) 1-2cos^2(y)/1-2cos(y)sin(y)=sin(y)+cos(y)/sin(y)-cos(y)

    please help or give some hints ,thank you .
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  2. #2
    MHF Contributor Calculus26's Avatar
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    For # 2 multiply the top and bottom by sin(y) -cos(y)

    expand and use sin^2(y) +cos^2(y) =1

    For # 1 I'm not sure it is an identity since the right hand side

    is csc^2(x) = cot^2(x) + 1 and if you expand the left hand side you get

    cot^2(x) +2cot(x) +1 -cot(x) = cot^2(x) +cot(x) +1
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  3. #3
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    Hello, starlet19!

    The first one is not an identity . . . There must be a typo.


    2)\;\;\frac{1-2\cos^2\!y}{1-2\cos y\sin y} \:=\:\frac{\sin y+\cos y}{\sin y-\cos y}
    Multiply the right side by \frac{\sin y - \cos y}{\sin y - \cos y}

    \frac{\sin y + \cos y}{\sin y - \cos y}\cdot\frac{\sin y - \cos y}{\sin y - \cos y} \;=\;\frac{\sin^2\!y - \cos^2\!y}{\sin^2\!y - 2\cos y\sin y + \cos^2\!y}

    . . =\;\frac{(1-\cos^2\!y) - \cos^2\!y}{\underbrace{\sin^2\!y + \cos^2\!y}_{\text{This is 1}}\: -\: 2\cos y\sin y}  \;=\;\frac{1-2\cos^2\!y}{1 - 2\cos y\sin y} \quad\hdots\quad\text{ta-}DAA!



    Ahhh, too fast for me, Calculus26!
    .
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