If $\tan \theta = - \frac{1}{{\sqrt 3 }}$ and $\sin \theta = \frac{1}{2}$, $\cos \theta = - \frac{{\sqrt 3 }}{2}$, then the principal value of $\theta $ will be
$\frac{\pi }{6}$
$\frac{{5\pi }}{6}$
$\frac{{7\pi }}{6}$
$ - \frac{\pi }{6}$
Solve $\tan 2 x=-\cot \left(x+\frac{\pi}{3}\right)$
The solution of the equation $\sec \theta - {\rm{cosec}}\theta = \frac{4}{3}$ is
For which value of $x$ ; $cosx > sinx,$ where $x\, \in \,\,\left( {\frac{\pi }{2}\,,\,\frac{{3\pi }}{2}} \right)$
If $\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$ is the solution of $4 \cos \theta+5 \sin \theta=1$, then the value of $\tan \alpha$ is
Solve $2 \cos ^{2} x+3 \sin x=0$