In a triangle ABC , let AB = 23 , BC = 3 and CA = 4 . Then the...

$\dfrac{cotA + cotC}{cotB} = \dfrac{\dfrac{cosA}{sinA} + \dfrac{cosC}{sinC}}{\dfrac{cosB}{sinB}}$

$= \dfrac{cosAsinC + cosCsinA}{sinAsinC} \cdot \dfrac{sinB}{cosB}$

$= \dfrac{\sin ( A + C )}{sinAsinC} \cdot \dfrac{sinB}{cosB}$

$= \dfrac{\sin ( \pi - B )}{sinAsinC} \cdot \dfrac{sinB}{cosB}$

$= \dfrac{\sin^{2}B}{sinAsinCcosB}$

Let, $AB = c, BC = a, CA = b$

By using sine rule, we get

$= \dfrac{{b}^{2}}{accosB}$

$= \dfrac{{b}^{2}}{ac\left(\dfrac{{a}^{2} + {c}^{2} - {b}^{2}}{2ac}\right)}$

$= \dfrac{2{b}^{2}}{{a}^{2} + {c}^{2} - {b}^{2}}$

$= \dfrac{32}{9 + 23 - 16}$

$= \dfrac{32}{16} = 2$