Question

If $\sqrt{3}\cos \theta -\sin \theta$ is positive,then $\theta$ lies between...

Answer 1

Consider the given expression.

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

Since, $\sqrt{3}\cos \theta -\sin \theta >0$

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

$\sqrt{3}>\tan \theta$

$\tan \theta <\sqrt{3}$

$\tan \theta <\tan \frac{\pi }{3}$

$\theta <\frac{\pi }{3}$

Hence, $-\infty <\theta <\frac{\pi }{3}$ is required answer.