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Equality on a triangle

Let ABC be a triangle. Show that

    \[\frac{1-\cos \hat{B}}{\sin \hat{B}} \cdot \frac{1-\cos \hat{\Gamma}}{\sin \hat{\Gamma}} = 1 - \frac{2a}{a+b+c}\]

Solution

Let s denote the semiperimeter of the triangle. On account of the well known relations,

(1)   \begin{equation*}\frac{1-\cos \hat{B}}{\sin \hat{B}} = \tan \frac{\hat{B}}{2} = \sqrt{\frac{\left ( s-a \right )\left ( s-c \right )}{s\left ( s-b \right )}} \end{equation*}

(2)   \begin{equation*}\frac{1-\cos \hat{\Gamma}}{\sin \hat{\Gamma}} = \tan \frac{\hat{\Gamma}}{2} = \sqrt{\frac{\left ( s-a \right )\left ( s-b \right )}{s\left ( s-c \right )}}\end{equation*}

we have:

    \begin{align*} \frac{1-\cos \hat{B}}{\sin \hat{B}} \cdot \frac{1-\cos \hat{\Gamma}}{\sin \hat{\Gamma}} &= \tan \frac{\hat{B}}{2} \cdot \tan \frac{\hat{\Gamma}}{2} \\ &=\sqrt{\frac{\left ( s-a \right )\left ( s-c \right )}{s\left ( s-b \right )}} \cdot \sqrt{\frac{\left ( s-a \right )\left ( s-b \right )}{s\left ( s-c \right )}} \\ &=\frac{s-a}{s} \\ &= 1- \frac{2a}{a+b+c} \end{align*}

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