In any $\Delta ABC$, prove that
$\dfrac{\left( a-b \right)}{c}\cos \dfrac{C}{2}=\sin \left( \dfrac{A-B}{2} \right)$

Hint: Try to simplify the left-hand side of the equation given in the question by the application of the sine rule of a triangle followed by the use of the formula of sin2A and the formula of (sinA-sinB).

Complete step-by-step answer:
Before starting with the solution, let us draw a diagram for better visualisation.

Now starting with the left-hand side of the equation that is given in the question.
We know, according to the sine rule of the triangle: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$ and in other terms, it can be written as:
$\begin{align}
  & a=k\sin A \\
 & b=k\sin B \\
 & c=k\sin C \\
\end{align}$
So, applying this to our expression, we get
$\dfrac{\left( a-b \right)}{c}\cos \dfrac{C}{2}$
$=\dfrac{k\sin A-k\sin B}{k\sin C}\times \cos \dfrac{C}{2}$
Now we will take $k$ common from all the terms. On doing so, we get
$=\dfrac{\left( \sin A-\sin B \right)}{\sin C}\times \cos \dfrac{C}{2}$
To further simplify the expression, we use the formula $\sin 2X=2\sin X\cos X$ .
$=\dfrac{\left( \sin A-\sin B \right)}{2\cos \dfrac{C}{2}\sin \dfrac{C}{2}}\times \cos \dfrac{C}{2}$
$=\dfrac{\left( \sin A-\sin B \right)}{2\sin \dfrac{C}{2}}$
Now we know that $\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ . On using this in our expression, we get
$=\dfrac{\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)}{\sin \dfrac{C}{2}}$
Now as ABC is a triangle, we can say:
$\angle A+\angle B+\angle C=180{}^\circ $
$\Rightarrow \angle A+\angle B=180{}^\circ -\angle C$
So, substituting the value of A+B in our expression. On doing so, we get
$=\dfrac{\cos \left( \dfrac{180{}^\circ -C}{2} \right)\sin \left( \dfrac{A-B}{2} \right)}{\sin \dfrac{C}{2}}$
$=\dfrac{\cos \left( 90{}^\circ -\dfrac{C}{2} \right)\sin \left( \dfrac{A-B}{2} \right)}{\sin \dfrac{C}{2}}$
We know $\cos \left( 90{}^\circ -X \right)=\sin X$ . Using this in our expression, we get
$=\dfrac{sin\dfrac{C}{2}\sin \left( \dfrac{A-B}{2} \right)}{\sin \dfrac{C}{2}}$
$=\sin \left( \dfrac{A-B}{2} \right)$
The left-hand side of the equation given in the question is equal to the right-hand side of the equation. Hence, we can say that we have proved the equation given in the question.

Note: Be careful about the calculation and the signs while opening the brackets. Also, you need to learn the sine rule and the cosine rule as they are used very often. The k in the sine rule is twice the radius of the circumcircle of the triangle, i.e., sine rule can also be written as $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k=2R=\dfrac{abc}{2\Delta }$ , where $\Delta $ represents the area of the triangle.