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Math Help - Analytic Trig Identities Problem. Need help asap!

  1. #1
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    Analytic Trig Identities Problem. Need help asap!

    The problem is..... (tan[theta] + cot[theta])^2=sec^2[theta] + scs^2[theta]
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  2. #2
    Junior Member sweet's Avatar
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    Wink

    Quote Originally Posted by kennysbaby28
    The problem is..... (tan[theta] + cot[theta])^2=sec^2[theta] + scs^2[theta]

    solved....
    (tan[theta] + cot[theta])^2=
    =(tan[theta])^2 +2(tan[theta]) ( cot[theta]) +(cot[theta])^2
    =(tan[theta])^2 +2(sin[theta]/cos[theta])(cos[theta]/sine[theta]+(cot[theta])^2
    =(tan[theta])^2 +2+(cot[theta])^2
    ={(tan[theta])^2 +1}+{(cot[theta])^2+1}
    an we know
    (tan[theta])^2 +1=sec^2[theta] and
    (cot[theta])^2 +1=csc^2[theta]

    then

    (tan[theta] + cot[theta])^2=sec^2[theta] + csc^2[theta]
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  3. #3
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    Question

    I'm not sure on how you got what you did and I'm not doubting you. It's just hard to understand the way it's written out. you have two sides to the problem so what goes on what side? Thanks for helping me!
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by kennysbaby28
    The problem is..... (tan[theta] + cot[theta])^2=sec^2[theta] + scs^2[theta]
    Please do not post the same question twice in different fora.

    RonL
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  5. #5
    Junior Member sweet's Avatar
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    Quote Originally Posted by kennysbaby28
    I'm not sure on how you got what you did and I'm not doubting you. It's just hard to understand the way it's written out. you have two sides to the problem so what goes on what side? Thanks for helping me!
    no prblem
    u take R.H.s
    <br />
(tan[theta] + cot[theta])^2=<br />
    <br />
=(tan[theta])^2 +2(tan[theta]) ( cot[theta]) +(cot[theta])^2<br />
    <br />
=(tan[theta])^2+<br />
    <br />
                     +2(sin[theta]/cos[theta])(cos[theta]/sine[theta]+<br />
    <br />
                                                                            +(cot[theta])^2<br />
    and know u have sin[theta] in numerator and sin[theta] in denominator also u have cos[theta] in numerator and cos[theta] in denominator
    then they drop so the eq be come
    <br />
=(tan[theta])^2 +2+(cot[theta])^2<br />
    <br />
={(tan[theta])^2 +1}+{(cot[theta])^2+1}<br />
    and we know
    <br />
(tan[theta])^2 +1=sec^2[theta]<br />

    and
    <br />
(cot[theta])^2 +1=csc^2[theta]<br />
    then
    <br />
(tan[theta] + cot[theta])^2=sec^2[theta] + csc^2[theta]<br />
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  6. #6
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    Hello, kennysbaby28!

    (\tan\theta + \cot\theta)^2 \:= \:\sec^2\theta + \csc^2\theta
    Expand the left side: \tan^2\theta + 2\tan\theta\cot\theta + \cot^2\theta

    Since \tan\theta\cdot\cot\theta\:=\:1, we have: \tan^2\theta + 2 + \cot^2\theta


    Since \tan^2\theta\:=\:\sec^2\theta - 1 and \cot^2\theta \:= \:\csc^2\theta - 1

    . . we have: (\sec^2\theta - 1) + 2 + (\csc^2\theta - 1) \;= \;\sec^2\theta + \csc^2\theta
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  7. #7
    Super Member malaygoel's Avatar
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    i will use tanx and cotx
    firstly, change tanx and cotx in sinx and cosx.
    add both of them
    you will get 1/(sinx*cosx)
    square the expression
    change 1 into sinx^2 +cosx^2
    you got the answer
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  8. #8
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    (\tan \theta + \cot \theta)^2 = \sec^2 \theta + \csc^2 \theta


    \tan^2 \theta + 2 \tan \theta \cot \theta + \cot^2 \theta = \sec^2 \theta + \csc^2 \theta


    \tan^2 \theta + 2 + \cot^2 \theta = \sec^2 \theta + \csc^2 \theta


    \frac{\sin^2 \theta}{\cos^2 \theta} + 2 + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}


    \frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} + 2 = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}


    \frac{2 \sin^2 \theta \cos^2 \theta + \sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta}


    2 \sin^2 \theta \cos^2 \theta + \sin^4 \theta + \cos^4 \theta = 1

    (\sin^2 \theta + \cos^2 \theta)^2 = 1

    (1)^2 = 1
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