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Math Help - proving trig identities

  1. #1
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    proving trig identities

     cot(A+B) = \frac{cotAcotB-1}{cotA+cotB}

    so far I have done;
     \frac{ \frac{1}{tanA} \times \frac{1}{tanB} - 1}{ \frac{1}{tanA} + \frac{1}{tanB}}

    which after simplifying I get;  \frac{1-tanAtanB}{tanA + tanB}

    I am stuck here, if anyone could direct me from here, thanks.
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  2. #2
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    How does your result relate to the tangent identity they gave you?
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  3. #3
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    Quote Originally Posted by Tweety View Post
     cot(A+B) = \frac{cotAcotB-1}{cotA+cotB}

    so far I have done;
     \frac{ \frac{1}{tanA} \times \frac{1}{tanB} - 1}{ \frac{1}{tanA} + \frac{1}{tanB}}

    which after simplifying I get;  \frac{1-tanAtanB}{tanA + tanB}

    I am stuck here, if anyone could direct me from here, thanks.

     \frac{1-\tan{A}\tan{B}}{\tan{A} + \tan{B}} = \frac{1}{\tan(A+B)} =  \cot(A+B)
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  4. #4
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    Quote Originally Posted by skeeter View Post
     \frac{1-\tan{A}\tan{B}}{\tan{A} + \tan{B}} = \frac{1}{\tan(A+B)} =  \cot(A+B)

    thanks,

    But I dont understand how you get  \frac{1}{\tan(A+B)}

    from  \frac{1-\tan{A}\tan{B}}{\tan{A} + \tan{B}}

    Is there some rule I am suppose to know? I understand that the denominator was factored, but what happened to  1-tanA tanB ? How does it become 1?
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  5. #5
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    Quote Originally Posted by Tweety View Post
    Is there some rule I am suppose to know?
    Yes: the tangent identity mentioned earlier. Check your chart or table of identities....
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  6. #6
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    Quote Originally Posted by stapel View Post
    Yes: the tangent identity mentioned earlier. Check your chart or table of identities....
    Oh I see what you mean, its this identity  tan(A + B) = \frac{tanA + tanB}{1-tanAtanB} but 'flipped' around?
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