In a triangle ABC, $a-2b+c=0$. The value of $\cot \left( \dfrac{A}{2} \right)\cot \left( \dfrac{C}{2} \right)$ is
(a) $\dfrac{9}{2}$
(b) $3$
(c) $\dfrac{3}{2}$
(d) $1$

Hint:Write $\cot \theta $ in the form of $\dfrac{\cos \theta }{\sin \theta }$. Find the relation between the angles A and C of the triangle, using the given information. Convert the angles in their half angle form, by using the expansion formula of sum of two angles of sine and cosine, to get the answer.

Complete step-by-step answer:
We know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$. Therefore, writing the given expression in this form, we get,
$\cot \left( \dfrac{A}{2} \right)\cot \left( \dfrac{C}{2} \right)$$=\dfrac{\cos \left( \dfrac{A}{2} \right)}{\sin \left( \dfrac{A}{2} \right)}\times \dfrac{\cos \left( \dfrac{C}{2} \right)}{\sin \left( \dfrac{C}{2} \right)}=\dfrac{\cos \left( \dfrac{A}{2} \right)\cos \left( \dfrac{C}{2} \right)}{\sin \left( \dfrac{A}{2} \right)\sin \left( \dfrac{C}{2} \right)}.................(i)$
Now, we have been given that: $a-2b+c=0$.
The relation between any side of a triangle and sine of the angle lying opposite to it is given by: $a=2R\sin A$, where $a=$ side of the triangle, $A=$ angle opposite to that of side $a$, and $R=$ radius of the circumcircle of the triangle.
$\begin{align}
  & \because a-2b+c=0 \\
 & \therefore a+c=2b \\
\end{align}$
Now, converting the given relation into its sine form, we get,
$2R\sin A+2R\sin C=2\times 2R\sin B$
Removing the common term from both sides we get,
$\sin A+\sin C=2\operatorname{Sin}B$
We know that, in a triangle: $A+B+C={{180}^{\circ }}$. Therefore,$B={{180}^{\circ }}-(A+C)$
$\Rightarrow \sin A+\sin C=2\sin ({{180}^{\circ }}-(A+C))$
We know that, $\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta $. Therefore,
$\sin A+\sin C=2\sin (A+C)$
Now, applying the formula $\sin a+\sin b=2\sin \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$ in the L.H.S and $\sin 2\theta =2\sin \theta \cos \theta $ i.e converting into half angle form $\sin \theta =2\sin (\dfrac{\theta}{2}) \cos (\dfrac{\theta}{2}) $ in the R.H.S, we get,
$2\sin \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\times 2\sin \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A+C}{2} \right)$
Cancelling the common terms we get,
$\cos \left( \dfrac{A-C}{2} \right)=2\cos \left( \dfrac{A+C}{2} \right)$
Applying the formula, $\cos (a\pm b)=\cos a\cos b\mp \sin a\sin b$, we have,
$\begin{align}
  & \cos \left( \dfrac{A}{2} \right)\cos \left( \dfrac{C}{2} \right)+\sin \left( \dfrac{A}{2} \right)\sin \left( \dfrac{C}{2} \right)=2\left( \cos \left( \dfrac{A}{2} \right)\cos \left( \dfrac{C}{2} \right)-\sin \left( \dfrac{A}{2} \right)\sin \left( \dfrac{C}{2} \right) \right) \\
 & \Rightarrow \cos \left( \dfrac{A}{2} \right)\cos \left( \dfrac{C}{2} \right)+\sin \left( \dfrac{A}{2} \right)\sin \left( \dfrac{C}{2} \right)=2\cos \left( \dfrac{A}{2} \right)\cos \left( \dfrac{C}{2} \right)-2\sin \left( \dfrac{A}{2} \right)\sin \left( \dfrac{C}{2} \right) \\
 & \Rightarrow \cos \left( \dfrac{A}{2} \right)\cos \left( \dfrac{C}{2} \right)=3\sin \left( \dfrac{A}{2} \right)\sin \left( \dfrac{C}{2} \right) \\
 & \Rightarrow \dfrac{\cos \left( \dfrac{A}{2} \right)\cos \left( \dfrac{C}{2} \right)}{\sin \left( \dfrac{A}{2} \right)\sin \left( \dfrac{C}{2} \right)}=\cot \left( \dfrac{A}{2} \right)\cot \left( \dfrac{C}{2} \right)=3 \\
\end{align}$
Therefore, the value of the expression (i) is 3.
Hence, option (b) is the correct answer.

Note: While solving the problems of ‘properties of a triangle’ always keep all the formulas in mind. Here, in the above solution we have changed the term that contains angle B into the term that contains angles A and C because the expression $\cot \left( \dfrac{A}{2} \right)\cot \left( \dfrac{C}{2} \right)$ does not contain angle B.Students should remember all the trigonometric identities and formulas for solving these types of questions and also the relation between any side of a triangle and sine of the angle lying opposite to it i.e $a=2R\sin A$ , $b=2R\sin B$, $c=2R\sin C$.