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Thread: prove that

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

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

    $\displaystyle b,c$ being sides of a triangle
    $\displaystyle A,B,C$ 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 $\displaystyle b^2(cotA + cotB) = c^2(cotA + cotC)$

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

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

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

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

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

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

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