Prove that; $$\dfrac{\cos \left( \pi -A\right) \cdot \cot \left( \dfrac{\pi }{2} +A\right) \cdot \cos \left( -A\right) }{\tan \left( \pi +A\right) \cdot \tan \left( \dfrac{3\pi }{2} +A\right) \cdot \sin \left( 2\pi -A\right) } =\cos \left(A\right) $$.

Hint: In this question a trigonometric equation is given, where we have to prove that LHS = RHS. So to find the solution we need to solve the left hand side of the equation. So while solving this we need some formulas, i.e,
1) $$\cos \left( \pi -\theta \right) =-\cos \theta$$
2) $$\cot \left( \dfrac{\pi }{2} +\theta \right) =-\tan \theta$$
3) $$\tan \left( \pi +\theta \right) =\tan \theta$$
4) $$\tan \left( \dfrac{3\pi }{2} +\theta \right) =-\cot \theta$$
5) $$\sin \left( 2\pi -\theta \right) =-\sin \theta$$
Complete step-by-step solution:
LHS,
$$\dfrac{\cos \left( \pi -A\right) \cdot \cot \left( \dfrac{\pi }{2} +A\right) \cdot \cos \left( -A\right) }{\tan \left( \pi +A\right) \cdot \tan \left( \dfrac{3\pi }{2} +A\right) \cdot \sin \left( 2\pi -A\right) }$$
$$= \dfrac{(-\cos A)\cdot (-\tan A)\cdot \cos A}{\tan A\cdot (-\cot A)\cdot(- \sin A)}$$[by using the above trigonometric formula]
$$= \dfrac{\cos A\cdot \tan A\cdot \cos A}{\tan A\cdot \cot A\cdot \sin A}$$
$$= \dfrac{\cos^{2} A}{\cot A\cdot \sin A}$$
$$= \dfrac{\cos^{2} A}{\left( \dfrac{\cos A}{\sin A} \right) \cdot \sin A}$$
$$= \dfrac{\cos^{2} A\cdot \sin A}{\cos A\cdot \sin A}$$=$$\cos \left( A\right) $$
$$= \cos A$$=RHS
Hence proved.
Note: While solving this type of question you need to know that basic formulas of trigonometry and also you have to know how the trigonometric functions changes with angles and the behaviour of a function in different angles, like the value of $\cos \left( \pi -\theta \right) $ gives $-\cos \theta$ but the value of $\sin \left( \pi -\theta \right) $ gives $\sin \theta$.