In a triangle ABC,cot A+cot B+cot C= | vedclass

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(C) Let a triangle $ABC$ have sides $a, b, c$ and area $\Delta$.We know that the area $\Delta = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B = \frac{1}

Vedclass · Accepted Answer

$\frac{a^2+b^2+c^2}{4 \Delta}$

Vedclass · Answer

(C) Let a triangle $ABC$ have sides $a, b, c$ and area $\Delta$.We know that the area $\Delta = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B = \frac{1}{2}ab \sin C$.From the cosine rule,we have $a^2 = b^2+c^2-2bc \cos A$,$b^2 = a^2+c^2-2ac \cos B$,and $c^2 = a^2+b^2-2ab \cos C$.Adding these three equations gives $a^2+b^2+c^2 = 2(a^2+b^2+c^2) - 2(bc \cos A + ac \cos B + ab \cos C)$.Rearranging,we get $a^2+b^2+c^2 = 2(bc \cos A + ac \cos B + ab \cos C)$.Since $\Delta = \frac{1}{2}bc \sin A$,we have $bc = \frac{2\Delta}{\sin A}$. Similarly,$ac = \frac{2\Delta}{\sin B}$ and $ab = \frac{2\Delta}{\sin C}$.Substituting these into the equation:$a^2+b^2+c^2 = 2 \left( \frac{2\Delta}{\sin A} \cos A + \frac{2\Delta}{\sin B} \cos B + \frac{2\Delta}{\sin C} \cos C \right)$$a^2+b^2+c^2 = 4\Delta (\cot A + \cot B + \cot C)$Therefore,$\cot A + \cot B + \cot C = \frac{a^2+b^2+c^2}{4\Delta}$.

In a $\triangle ABC$,$\cot A+\cot B+\cot C=$

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