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Math Help - Proving Problems

  1. #1
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    Exclamation Proving Problems

    Have an assignment due in the near future, and these four are giving me a hassle to prove, any help or advice would be appreciated.

    Prove
    <br />
\frac{1+tan{x}}{sinx}= cscxsecx<br />

     <br />
tanx+cotx=\frac{csc^2{x}}{cotx}<br />
    Last edited by nightrider456; January 17th 2010 at 12:15 PM.
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  2. #2
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    Quote Originally Posted by nightrider456 View Post
    Have an assignment due in the near future, and these four are giving me a hassle to prove, any help or advice would be appreciated.

    Prove
    <br />
\frac{1+sin{x}}{sinx}= cscxsecx<br />

     <br />
tanx+cotx=\frac{csc^2{x}}{cotx}<br />

     <br />
\frac{Cos2{x}}{sinx}= \frac{Cot^2{x}-1}{csc{x}}<br />

     <br />
Sin2x(tanx+cotx)=2<br />
    \tan{x} + \cot{x} = \frac{\sin{x}}{\cos{x}} + \frac{\cos{x}}{\sin{x}}

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

     = \frac{1}{\sin{x}\cos{x}}

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

     = \frac{1}{\sin^2{x}}\cdot \frac{\sin{x}}{\cos{x}}

     = \csc^2{x}\tan{x}

     = \frac{\csc^2{x}}{\cot{x}}.
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  3. #3
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    Hello, nightrider456!

    The first one is not an identity.
    Please check for typos.


     \frac{\cos2{x}}{\sin x}\:=\: \frac{\cot^2\!x-1}{\csc x}

    Start with the left side: . \frac{\cos2x}{\sin x} \;=\;\frac{\cos^2\!x-\sin^2\!x}{\sin x}


    Divide top and bottom by \sin^2\!x:\;\;\frac{\dfrac{\cos^2x}{\sin^2x} - \dfrac{\sin^2x}{\sin^2x}}{\dfrac{\sin x}{\sin^2x}}


    And we have: . \frac{\left(\dfrac{\cos x}{\sin x}\right)^2 - 1}{\dfrac{1}{\sin x}} \;=\;\frac{\cot^2\!x - 1}{\csc x}

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    Sorry, fixed it now thanks for having a look
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  5. #5
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    Quote Originally Posted by nightrider456 View Post
    Have an assignment due in the near future, and these four are giving me a hassle to prove, any help or advice would be appreciated.

    Prove
    <br />
\frac{1+tan{x}}{sinx}= cscxsecx<br />

     <br />
tanx+cotx=\frac{csc^2{x}}{cotx}<br />
    \frac{1 + \tan{x}}{\sin{x}} = \frac{1}{\sin{x}} + \frac{\tan{x}}{\sin{x}}

     = \csc{x} + \frac{\frac{\sin{x}}{\cos{x}}}{\sin{x}}

     = \csc{x} + \frac{1}{\cos{x}}

     = \csc{x} + \sec{x}...


    Are you sure you weren't supposed to prove it's equal to \csc{x} + \sec{x} instead of \csc{x}\sec{x}?
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