If \[\sin A = \sin B\] and $\cos A = \cos B,$ then
A. $\sin \left( {\dfrac{{A - B}}{2}} \right) = 0$
B. $\sin \left( {\dfrac{{A + B}}{2}} \right) = 0$
C. $\cos \left( {\dfrac{{A - B}}{2}} \right) = 0$
D. $\cos \left( {A + B} \right) = 0$

Hint: In order to solve this type of question, first we will simplify both the given equations using the suitable trigonometric identities. Then, we will consider another suitable trigonometric identity. Next, we will substitute the values in it and simplify it to get the correct answer.

Formula used:
$\left[ {\because \sin 2\theta = 2\sin \theta \cos \theta } \right]$
$\left[ {\because \cos 2\theta = 2{{\cos }^2}\theta - 1} \right]$
$\sin \left( {\dfrac{{A - B}}{2}} \right) = \sin \dfrac{A}{2}\cos \dfrac{B}{2} - \cos \dfrac{A}{2}\sin \dfrac{B}{2}$

Complete step by step solution:
We are given that,
\[\sin A = \sin B\] ………………..equation$\left( 1 \right)$
$\cos A = \cos B$ ………………..equation$\left( 2 \right)$
Simplifying equation$\left( 1 \right)$,
\[2\sin \dfrac{A}{2}\cos \dfrac{A}{2} = 2\sin \dfrac{B}{2}\cos \dfrac{B}{2}\] $\left[ {\because \sin 2\theta = 2\sin \theta \cos \theta } \right]$
\[\sin \dfrac{A}{2}\cos \dfrac{A}{2} = \sin \dfrac{B}{2}\cos \dfrac{B}{2}\] ………………..equation$\left( 3 \right)$
Simplifying equation$\left( 2 \right)$,
$2{\cos ^2}\dfrac{A}{2} - 1 = 2{\cos ^2}\dfrac{B}{2} - 1$ $\left[ {\because \cos 2\theta = 2{{\cos }^2}\theta - 1} \right]$
$\cos \dfrac{A}{2} = \cos \dfrac{B}{2}$ ………………..equation$\left( 4 \right)$
From equation $\left( 3 \right)$ and $\left( 4 \right)$ we get,
\[\sin \dfrac{A}{2} = \sin \dfrac{B}{2}\] ………………..equation$\left( 5 \right)$
Consider,
$\sin \left( {\dfrac{{A - B}}{2}} \right) = \sin \dfrac{A}{2}\cos \dfrac{B}{2} - \cos \dfrac{A}{2}\sin \dfrac{B}{2}$
Using equation $\left( 4 \right)$ and $\left( 5 \right)$ in the above equation we get,
$\sin \left( {\dfrac{{A - B}}{2}} \right) = 0$
$\therefore $ The correct option is A.

Note: To solve this question one has to choose the suitable trigonometric identities and be very sure while simplifying them. This type of question requires the use of correct application of trigonometric rules to get the correct answer. We should focus on options to choose our suitable identity for the simplification.