Question

In a triangle $ABC$, let $AB=\sqrt{23},BC=3$ and $CA=4$. Then the value of $\frac{\cot A+\cot C}{\cot B}$ is ______.

Answer 1

Answer: 2

$AB=\sqrt{23}=C$
$BC=3=a$
$CA=4=b$
$\frac{\cot A+\cot C}{\cot B};\frac{\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}}{\frac{\cos B}{\sin B}}$
$=\frac{\cos A\sin C+\cos C\sin A}{\sin A\cdot \sin C\cdot \frac{\cos B}{\sin B}}=\frac{\sin (A+C)\cdot \sin B}{\sin A\cdot \sin C\cdot \cos B}$
$=\frac{\sin B\cdot \sin B}{\sin A\cdot \sin C\cdot \cos B}$
$=\frac{b^2}{ac\cdot \frac{\left(a^2+c^2-b^2\right)}{2ac}}=\frac{2b^2}{a^2+c^2-b^2}$
$=\frac{2\times 16}{9+23-16}=\frac{2\times 16}{32-16}$
$=\frac{2\times 16}{16}=2$