The \[\tan \theta .\sin \left( {\dfrac{\pi }{2} + \theta } \right).\cos \left( {\dfrac{\pi }{2} - \theta } \right)\]=?
A. 1
B. 0
C. \[\dfrac{1}{{\sqrt 2 }}\]
D. None of these

Hint: Given is the combination of three trigonometric functions which are interrelated. Like tan is the ratio of sin and cos. Thus we will change the ratios as per the requirement. Then we will cancel the functions that can be and then finalize the answer.

Complete step by step answer:
Given is the sum,
\[\tan \theta .\sin \left( {\dfrac{\pi }{2} + \theta } \right).\cos \left( {\dfrac{\pi }{2} - \theta } \right)\]
We know that,
\[\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta \] and \[\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \]
\[ \Rightarrow \tan \theta .\cos \theta .\sin \theta \]
We know that, \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}.\cos \theta .\sin \theta \]
We can cancel the \[\cos \theta \] function,
\[\therefore \tan \theta = {\sin ^2}\theta \]
But the options are not having any of the answers above.

So option D is the correct answer.

Note:This is very simple problem to solve only the thing that is to be noted and to be taken care of is the added angle formulas of \[\sin \left( {\dfrac{\pi }{2} + \theta } \right)\,and\,\cos \left( {\dfrac{\pi }{2} - \theta } \right)\]. Check whether the angle is added or subtracted.According to that, we will take the respective angle.