Prove that $$\cos \left( {\dfrac{\pi }{6} - A} \right) \cdot \cos \left( {\dfrac{\pi }{3} + B} \right) - \sin \left( {\dfrac{\pi }{6} - A} \right) \cdot \sin \left( {\dfrac{\pi }{3} + B} \right) = \cos \left( {A - B} \right)$$

Hint: Here in this question, we have to prove the given trigonometric function by showing the left hand side is equal to the right hand side (i.e., $$L.H.S = R.H.S$$). To solve this, we have to consider L.H.S separately and simplify by using a formula of compound angle of trigonometric ratios $$\cos \left( {A + B} \right) = \cos A \cdot \cos B - \sin A \cdot \sin B$$ to get the required RHS.

Complete step by step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function Also called circular function.
Consider the given question:
Prove that
$$\cos \left( {\dfrac{\pi }{6} - A} \right) \cdot \cos \left( {\dfrac{\pi }{3} + B} \right) - \sin \left( {\dfrac{\pi }{6} - A} \right) \cdot \sin \left( {\dfrac{\pi }{3} + B} \right) = \sin \left( {A - B} \right)$$ --------(1)
Consider Left hand side of equation (1) (L.H.S)
$$ \Rightarrow \,\,\,\cos \left( {\dfrac{\pi }{6} - A} \right) \cdot \cos \left( {\dfrac{\pi }{3} + B} \right) - \sin \left( {\dfrac{\pi }{6} - A} \right) \cdot \sin \left( {\dfrac{\pi }{3} + B} \right)$$ ----(2)
Let us by the compound angles of trigonometric ratios:
The sum identity of cosine ratio is:
$$\cos \left( {A + B} \right) = \cos A \cdot \cos B - \sin A \cdot \sin B$$
On comparing the equation (2) with cosine sum identity
Where, $$A = \left( {\dfrac{\pi }{6} - A} \right)$$ and $$B = \left( {\dfrac{\pi }{3} + B} \right)$$, then
By cosine sum identity equation (2) becomes
$$= \cos \left( {\left( {\dfrac{\pi }{6} - A} \right) + \left( {\dfrac{\pi }{3} + B} \right)} \right)$$
By using a sign conversion
$$ = \cos \left( {\dfrac{\pi }{6} - A + \dfrac{\pi }{3} + B} \right)$$
$$=\cos \left( {\dfrac{\pi }{6} + \dfrac{\pi }{3} - \left( {A - B} \right)} \right)$$
Take, 6 as LCM between $$\dfrac{\pi }{6}$$ and $$\dfrac{\pi }{3}$$, then we have
$$=\cos \left( {\dfrac{{\pi + 2\pi }}{6} - \left( {A - B} \right)} \right)$$
$$= \cos \left( {\dfrac{{3\pi }}{6} - \left( {A - B} \right)} \right)$$
On simplification, we get
$$ = \cos \left( {\dfrac{\pi }{2} - \left( {A - B} \right)} \right)$$ ---- (3)
By the ASTC rule $$\left( {\dfrac{\pi }{2} - \theta } \right)$$ belongs to the first quadrant in that all six ratios are positive and
Let us by the complementary angles of trigonometric ratios:
The angle can be written as
$$\sin \left( {90 - \theta } \right) = \cos \theta $$
$$\cos \left( {90 - \theta } \right) = \sin \theta $$
Then equation (3) becomes
$$ \Rightarrow \,\,\,\sin \left( {A - B} \right)$$
$$ = RHS$$
$$\therefore \,\,LHS = \,\,RHS$$.
$$\therefore \,\,\,\,\cos \left( {\dfrac{\pi }{6} - A} \right) \cdot \cos \left( {\dfrac{\pi }{3} + B} \right) - \sin \left( {\dfrac{\pi }{6} - A} \right) \cdot \sin \left( {\dfrac{\pi }{3} + B} \right) = \sin \left( {A - B} \right)$$
Hence, proved.

Note:
When solving the trigonometry-based questions, we have to know the definitions of six trigonometric ratios. Remember, the formula of compound angles i.e.,
cosine sum identity: $$\cos \left( {A + B} \right) = \cos A \cdot \cos B - \sin A \cdot \sin B$$ and
cosine difference identity: $$\cos \left( {A - B} \right) = \cos A \cdot \cos B + \sin A \cdot \sin B$$
Sine sum identity: $$\sin \left( {A + B} \right) = \sin A \cdot \cos B + \cos A \cdot \sin B$$
Sine difference identity: $$\sin \left( {A - B} \right) = \sin A \cdot \cos B - \cos A \cdot \sin B$$
Remember, when the sum of two angles is $${90^ \circ }$$, then the angles are known as complementary angles at that time the ratios will change like $$\sin \leftrightarrow \cos $$, $$\sec \leftrightarrow cosec$$ and $$\tan \leftrightarrow \cot $$ then should know the some basic formulas of trigonometry like identities, double and half angle formulas, Product to Sum Formulas and Sum to Product Form.