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Question: The \[\tan \theta .\sin \left( {\dfrac{\pi }{2} + \theta } \right).\cos \left( {\dfrac{\pi }{2} - \t...

The tan⁡θ.sin⁡(π2+θ).cos⁡(π2−θ)\tan \theta .\sin \left( {\dfrac{\pi }{2} + \theta } \right).\cos \left( {\dfrac{\pi }{2} - \theta } \right)=?
A. 1
B. 0
C. 12\dfrac{1}{{\sqrt 2 }}
D. None of these

Explanation

Solution

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⁡θ.sin⁡(π2+θ).cos⁡(π2−θ)\tan \theta .\sin \left( {\dfrac{\pi }{2} + \theta } \right).\cos \left( {\dfrac{\pi }{2} - \theta } \right)
We know that,
sin⁡(π2+θ)=cos⁡θ\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta and cos⁡(π2−θ)=sin⁡θ\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta
⇒tan⁡θ.cos⁡θ.sin⁡θ\Rightarrow \tan \theta .\cos \theta .\sin \theta
We know that, tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
tan⁡θ=sin⁡θcos⁡θ.cos⁡θ.sin⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}.\cos \theta .\sin \theta
We can cancel the cos⁡θ\cos \theta function,
∴tan⁡θ=sin⁡2θ\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⁡(π2+θ) and cos⁡(π2−θ)\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.