Question

Find the smallest positive value of $\theta$ satisfying $\tan \theta =2\sin \theta$.

Answer 1

Since, $\frac{\tan \theta }{\sin \theta }=2\sin \theta$
$\Rightarrow \frac{\sin \theta }{\cos \theta }=2\sin \theta$ or $\sin \theta =2\sin \theta \cos \theta$
$\Rightarrow \sin \theta =0$ or $1-2\cos \theta =0$ i.e. $\cos \theta =\frac{1}{2}$
If $\sin \theta =0$ the least + ve value of $\theta =0^0$
If $\cos \theta =\frac{1}{2}$ the least + ve value of $\theta$ is ${60}^0$
$\therefore \theta =0^0$ is the least + ve value