Question

All possible values of $\theta \in \left[0,2\mathrm{\pi }\right]$ for which $\sin 2\theta +\tan 2\theta >0$ lie in

• A. $\left(0,\left(\frac{\pi }{2}\right)\right)\cup \left(\pi ,\frac{3\pi }{2}\right)$
• B. $\left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right)$
• C. $\left(0,\frac{\mathrm{\pi }}{2}\right)\cup \left(\frac{\mathrm{\pi }}{2}\frac{3\mathrm{\pi }}{4}\right)\cup \left(\mathrm{\pi },\frac{7\mathrm{\pi }}{6}\right)$
• D. $\left(0,\frac{\mathrm{\pi }}{4}\right)\cup \left(\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{4}\right)\cup \left(\frac{3\mathrm{\pi }}{2},\frac{11\mathrm{\pi }}{6}\right)$

Answer 1

The correct option is B

$\left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right)$

Explanation for the correct option:

Finding the value:

$\sin \left(2\theta \right)+\tan \left(2\theta \right)>0$

$$\begin{gathered} \Rightarrow \left(\sin \left(2\theta \right)\right)+\frac{\sin \left(2\theta \right)}{\cos \left(2\theta \right)}>0 \\ \Rightarrow \sin \left(2\theta \right)\frac{1+\cos 2\theta }{\cos \left(2\theta \right)}>0 \\ \Rightarrow \tan \left(2\theta \right)\left(2{\cos }^2\theta \right)>0 \\ Now,\tan \left(2\theta \right)>0 \\ \Rightarrow 2\theta =\left(0,\frac{\pi }{2}\right)\cup \left(\pi ,\frac{3\pi }{2}\right)\cup \left(2\pi ,\frac{5\pi }{2}\right)\cup \left(3\pi ,\frac{7\pi }{2}\right) \\ \Rightarrow \theta =\left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right) \end{gathered}$$

Hence, option(B) is correct.