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

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