Question

If $\sqrt{3}\sin \theta -\cos \theta =0$ and $0^{\circ }<\theta <{90}^{\circ }$, Find $\theta$.

Answer 1

We have,

$\sqrt{3}\sin \theta -\cos \theta =0$

$\sqrt{3}\sin \theta =\cos \theta$

$\sqrt{3}=\frac{\cos \theta }{\sin \theta }$

$\sqrt{3}=\cot \theta$

$\tan \theta =\frac{1}{\sqrt{3}}$

$\tan \theta =\tan \frac{\pi }{6}$

$\theta =n\pi +\frac{\pi }{6}$ where $n\epsilon Z$

Hence, this is the answer.