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Math Help - Identities

  1. #1
    Member purplec16's Avatar
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    Identities

    Can someone help me with these two problems
    sec \theta + csc \theta-cos\theta-sin\theta= sin\theta tan\theta+ cos\theta cot\theta

    and

    \frac{cot(-t)+tan(-t)}{cot t}= -sec^2 t
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  2. #2
    Senior Member Stroodle's Avatar
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    For the first one, you can take the RHS:

    sin\theta tan\theta+cos\theta cot\theta = sin\theta\times\frac{sin\theta}{cos\theta}+cos\the  ta\times\frac{cos\theta}{sin\theta}

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

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

    =\frac{1}{cos\theta}-cos\theta+\frac{1}{sin\theta}-sin\theta

    =sec\theta+csc\theta-cos\theta-sin\theta = LHS
    Last edited by Stroodle; March 9th 2010 at 05:56 PM.
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  3. #3
    Member purplec16's Avatar
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    Do you know how to do the other one?
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  4. #4
    Senior Member Stroodle's Avatar
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    For the second one you can take the LHS:

    \frac{cot(-t)+tan(-t)}{cot(t)}=\frac{-cot(t)-tan(t)}{cot(t)}

    =-1-\frac{\frac{sin(t)}{cos(t)}}{\frac{cos(t)}{sin(t)}  }

    =-1-\frac{sin^2(t)}{cos^2(t)}

    =-1-tan^2(t)

    =-sec^2(t)=RHS
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  5. #5
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    Quote Originally Posted by purplec16 View Post
    Can someone help me with these two problems
    sec \theta + csc \theta-cos\theta-sin\theta= sin\theta tan\theta+ cos\theta cot\theta

    and

    \frac{cot(-t)+tan(-t)}{cot t}= -sec^2 t
    \frac{\cot{(-t)} + \tan{(-t)}}{\cot{t}} = \frac{\frac{\cos{(-t)}}{\sin{(-t)}} + \frac{\sin{(-t)}}{\cos{(-t)}}}{\frac{\cos{t}}{\sin{t}}}

     = \frac{\frac{\cos{t}}{-\sin{t}} + \frac{-\sin{t}}{\cos{t}}}{\frac{\cos{t}}{\sin{t}}}

     = \frac{-\cot{t} - \tan{t}}{\cot{t}}

     = -1 - \frac{\tan{t}}{\cot{t}}

     = -1 - \frac{\tan{t}}{\frac{1}{\tan{t}}}

     = -1 - \tan^2{t}

     = -(1 + \tan^2{t})

     = -\sec^2{t}.
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  6. #6
    Member purplec16's Avatar
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    THANK YOU SO MUCH!!!!!!!!!!!!!!!!!
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