In a triangle $\mathrm{ABC}, l(\mathrm{AB})=\sqrt{23}$ units, $l(\mathrm{BC})=3$ units, $l(\mathrm{CA})=4$…
In a triangle $\mathrm{ABC}, l(\mathrm{AB})=\sqrt{23}$ units, $l(\mathrm{BC})=3$ units, $l(\mathrm{CA})=4$ units, then $\frac{\cot A+\cot C}{\cot B}$ is
1
2
4
8
Solution
$\begin{aligned} & \frac{\cot A+\cot C}{\cot B} \\ = & \frac{\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}}{\frac{\cos B}{\sin B}}\end{aligned}$
$=\frac{\frac{b^2+c^2-a^2}{2 b c \sin A}+\frac{a^2+b^2-c^2}{2 a b \sin C}}{\frac{c^2+a^2-b^2}{2 c a \sin B}}$
$=\frac{\frac{b^2+c^2-a^2}{2(2 \Delta)}+\frac{a^2+b^2-c^2}{2(2 \Delta)}}{\frac{c^2+a^2-b^2}{2(2 \Delta)}}$
$\begin{aligned} & =\frac{2 b^2}{c^2+a^2-b^2} \\ & =\frac{2(4)^2}{(\sqrt{23})^2+(3)^2-(4)^2} \\ & =\frac{32}{16} \\ & =2\end{aligned}$