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Math Help - tan(A/2) + cot9A/2) = 2cosecA

  1. #1
    Member Furyan's Avatar
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    tan(A/2) + cot(A/2) = 2cosecA

    Hello

    Could someone please help me start with proving the following identity:

    \tan \dfrac{A}{2} + \cot \dfrac{A}{2} \equiv 2\csc A

    I'm trying to show that the L.H.S = the R.H.S. I've tried finding the half angle formula for \tan \dfrac{A}{2}, but it looks like a quadratic. I have also tried writing the L.H.S as a single term. I'm totally stuck. A hint would be greatly appreciated.

    Thank you
    Last edited by Furyan; January 20th 2013 at 12:20 PM.
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  2. #2
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    Re: tan(A/2) + cot9A/2) = 2cosecA

    Quote Originally Posted by Furyan View Post
    Hello

    Could someone please help me start with proving the following identity:

    \tan \dfrac{A}{2} + \cot \dfrac{A}{2} \equiv 2\csc A

    To save notation let u=\frac{A}{2}.

    \frac{\sin(u)}{\cos(u)}+\frac{\cos(u)}{\sin(u)}= \frac{1}{\sin(u)\cos(u)}=\frac{2}{\sin(2u)}
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  3. #3
    Member Furyan's Avatar
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    Re: tan(A/2) + cot9A/2) = 2cosecA

    Hello Plato,

    Thank you so much. I clearly wasn't trying hard enough to write the L.H.S as a single term. I totally missed that.

    Thank you for all your help and patience, it's very much appreciated.
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  4. #4
    Member Furyan's Avatar
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    Re: tan(A/2) + cot9A/2) = 2cosecA

    ...and I actually understand the use of the double angle formula to equate the following.

     \frac{1}{\sin(u)\cos(u)}=\frac{2}{\sin(2u)}

    Edit: Understand might be an over statement, but I get it.
    Last edited by Furyan; January 20th 2013 at 12:56 PM.
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