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Math Help - problem with proving identities using basic identities

  1. #1
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    problem with proving identities using basic identities

    i have a problem with this:

    prove,
    a) \frac{1-tan^2\theta}{cot^2 \theta-1}=tan^2\theta


    b) (cos \theta+sin \theta)^2+(cos \theta-sin \theta)^2=2

    any help will appreciate!! thanks.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    a) \frac{1-tan^2\theta}{cot^2\theta-1} = \frac{1-\frac{sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta  }{sin^2\theta} - 1}

     = \frac{\frac{cos^2\theta - sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta - sin^2\theta}{sin^2\theta}}

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

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

    =tan^2\theta
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  3. #3
    MHF Contributor Unknown008's Avatar
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    b) (cos\theta + sin\theta)^2 + (cos\theta - sin\theta)^2 = cos^2 \theta + 2cos\theta sin\theta + sin^2\theta + cos^2\theta -2cos\theta sin\theta + sin^2\theta

    = cos^2 \theta + \cancel{2cos\theta sin\theta} + sin^2\theta + cos^2\theta \cancel{-2cos\theta sin\theta} + sin^2\theta

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

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

     = 2(1)

     = 2
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  4. #4
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    Quote Originally Posted by Unknown008 View Post
    a) \frac{1-tan^2\theta}{cot^2\theta-1} = \frac{1-\frac{sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta  }{sin^2\theta} - 1}

     = \frac{\frac{cos^2\theta - sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta - sin^2\theta}{sin^2\theta}}

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

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

    =tan^2\theta

    i want to know,how can \frac{cos^2\theta-sin^2\theta}{cos^2 \theta}=\frac{1}{cos^2\theta} ??
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  5. #5
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    Because \frac{\frac{\cos^2{\theta} - \sin^2{\theta}}{\cos^2{\theta}}}{\frac{\cos^2{\the  ta} - \sin^2{\theta}}{\sin^2{\theta}}} = \frac{\frac{1}{\cos^2{\theta}}}{\frac{1}{\sin^2{\t  heta}}}\left(\frac{\cos^2{\theta} - \sin^2{\theta}}{\cos^2{\theta} - \sin^2{\theta}}\right)

     =  \frac{\frac{1}{\cos^2{\theta}}}{\frac{1}{\sin^2{\t  heta}}}
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Ok, I'll break up the steps, but this is not what I meant.

     = \frac{\frac{cos^2\theta - sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta - sin^2\theta}{sin^2\theta}}

     = \frac{cos^2\theta - sin^2\theta}{cos^2\theta} \times \frac{sin^2\theta}{cos^2\theta - sin^2\theta}

    Then:

     = \frac{1}{cos^2\theta} \times \frac{sin^2\theta}{1}
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  7. #7
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    thanks very much both of you sir!!
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