If  $\frac{1}{6}\sin \theta ,\cos \theta ,\tan \theta$  are in G.P., then $\theta$ is equal to $(n\in Z)$

a,b, c are in GP then  b²=ac

${\cos }^2\theta =\frac{1}{6}\sin \theta \tan \theta$   or

$6{\cos }^3\theta ={\sin }^2\theta$

$(2\cos \theta -1)(3{\cos }^2\theta +2\cos \theta +1)=0$   or

$\cos \theta =\frac{1}{2}\Rightarrow \theta =2n\pi \pm \frac{\pi }{3},n\in Z$