Question

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

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

Answer 1

Answer: (D)

$\sin 2\theta +\tan 2\theta >0$
$\Rightarrow \sin 2\theta +\frac{\sin 2\theta }{\cos 2\theta }>0$
$\Rightarrow \sin 2\theta \frac{(\cos 2\theta +1)}{\cos 2\theta }>0\Rightarrow \tan 2\theta \left(2{\cos }^2\theta \right)>0$
Note $:\cos 2\theta \ne 0$
$\Rightarrow 1-2{\sin }^2\theta \ne 0\Rightarrow \sin \theta \ne \pm \frac{1}{\sqrt{2}}$
Now, $\tan 2\theta (1+\cos 2\theta )>0$
$\Rightarrow \tan 2\theta >0($ as $\cos 2\theta +1>0)$
$\Rightarrow 2\theta \in \left(0,\frac{\pi }{2}\right)\cup \left(\pi ,\frac{3\pi }{2}\right)\cup \left(2\pi ,\frac{5\pi }{2}\right)\cup \left(3\pi ,\frac{7\pi }{2}\right)$
$\Rightarrow \theta \in \left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right)$
As $\sin \theta \ne \pm \frac{1}{\sqrt{2}};$ which has been already considered