Question

Evaluate $\tan \theta \sin \left(\frac{\mathrm{\pi }}{2}+\theta \right)\cos \left(\frac{\mathrm{\pi }}{2}-\theta \right)$

• A. $1$
• B. $0$
• C. $\frac{1}{\sqrt{2}}$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for correct option:

Simplifying the given expression:

Given, $\mathrm{tan\theta sin}\left(\frac{\mathrm{\pi }}{2}+\mathrm{\theta }\right)\cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)$

We know that,

Using the above two identities in the given expression we get,

$$\begin{gathered} =\mathrm{tan\theta sin}\left(\frac{\mathrm{\pi }}{2}+\mathrm{\theta }\right)\cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right) \\ =\mathrm{tan\theta cos\theta sin\theta } \\ =\frac{\mathrm{sin\theta }}{\cos \mathrm{\theta }}\mathrm{cos\theta sin\theta } \\ ={\sin }^2\mathrm{\theta } \end{gathered}$$ $$\begin{gathered} [\because sin\left(\frac{\pi }{2}+\theta \right)=cos\theta ,\frac{sin\theta }{cos\theta }=tan\theta \\ cos\left(\frac{\pi }{2}-\theta \right)=sin\theta ] \end{gathered}$$

Hence, the correct answer is option (D).