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Math Help - triangles is tough

  1. #1
    Member grgrsanjay's Avatar
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    triangles is tough

    in a triangle ABC where a,b,c are the sides opposite to the angles A,B,C respectively
    AC is double of AB, then find the value of {cot\:\dfrac{A}{2} }{cot\:\dfrac{B-C}{2} }
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    Quote Originally Posted by grgrsanjay View Post
    in a triangle ABC where a,b,c are the sides opposite to the angles A,B,C respectively
    AC is double of AB, then find the value of {cot\:\dfrac{A}{2} }{cot\:\dfrac{B-C}{2} }
    {\dfrac{AB}{AC} = \dfrac{1}{2} = \dfrac{\sin{C}}{\sin{B}}

    = {\dfrac{(2+1)}{(2-1)} = \dfrac{(sinB +sinC)}{(sinB - sinC)}}

    = {\dfrac{2sin\dfrac{B+C}{2}cos\dfrac{B-C}{2}}{2cos\dfrac{B+C}{2}sin\dfrac{B-C}{2}}}

    Since (B+C)/2 = π/2 - A/2, sin[(B+C)/2] = cos(A/2) and cos[(B+C)/2] = sin(A/2).

    Substtitue these values in the above equation and simplify.
    Last edited by sa-ri-ga-ma; August 11th 2010 at 10:34 PM.
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