JEE Advanced 2021 · Paper 1, Mathematics question 18

Cotangents in terms of the sides. With area $\Delta$, $\cot A=\dfrac{b^2+c^2-a^2}{4\Delta}$, and similarly for $B$ and $C$. Therefore
$$\frac{\cot A+\cot C}{\cot B}=\frac{2b^2}{a^2+c^2-b^2}.$$

Substitute $a=BC=3$, $b=CA=4$, $c=AB=\sqrt{23}$:
$$\frac{32}{9+23-16}=2.$$