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Math Help - prove that

  1. #1
    Super Member bigwave's Avatar
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    Talking prove that

    prove that b^2(cotA + cotB) = c^2(cotA + cotC)

    b,c being sides of a triangle
    A,B,C being corresponding interior angles
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by bigwave View Post
    prove that b^2(cotA + cotB) = c^2(cotA + cotC)

    b,c being sides of a triangle
    A,B,C being corresponding interior angles
    \frac{cot A+cot C}{cot A+cot B}=\frac{\frac{cos A}{sin A}+\frac{cos C}{sin C}}{\frac{cos A}{sin A}+\frac{cos B}{sin B}}

    = \frac{sin (A+C)}{sin A sin C}*\frac{sin A sin B}{sin(A+B)}

    = \frac{sin (\pi-B)}{sin C}*\frac{sin B}{sin(\pi-C)}

    = \frac{sin B}{sin C}*\frac{sin B}{sin C}

    = \frac{sin^2B}{sin^2C}

    = \frac{b^2}{c^2}

    Thus, b^2(cotA + cotB) = c^2(cotA + cotC)
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