# Question 2 [JEE adv, 2021]

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

## Concepts

Use $$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$$ and $$\Delta = \dfrac{1}{2}bc\sin A$$

## Solution

#### Step 1

$\begin{eqnarray} \frac{\cot A + \cot C}{\cot B} &=& \frac{\frac{\cos A}{\sin A} + \frac{\cos C}{\sin C}}{\frac{\cos B}{\sin B}}\\ &=&\frac{\frac{b^2 + c^2 - a^2}{2bc\sin A} + \frac{a^2 + b^2 - c^2}{2ab\sin C}}{\frac{a^2 + c^2 - b^2}{2ac\sin B}}\\ &=&\frac{\frac{b^2 + c^2 - a^2}{4\Delta} + \frac{a^2 + b^2 - c^2}{4\Delta}}{\frac{a^2 + c^2 - b^2}{4\Delta}}\\ &=& \frac{2b^2}{a^2 + c^2 - b^2}\\ &=& \frac{2\times 4^2}{3^2 + 23 - 4^2} = 2 \end{eqnarray}$