If $\frac{\sin \mathrm{\theta }}{6},\cos \mathrm{\theta } \mathrm{and} \tan \mathrm{\theta } \mathrm{are}\mathrm{in} \mathrm{GP}\mathrm{, }$then the general value of $\mathrm{\theta }$ is 

Since,${\cos }^2\mathrm{\theta }=\frac{1}{6}\sin \mathrm{\theta }\cdot \tan \mathrm{\theta }$

$\Rightarrow 6{\cos }^3\mathrm{\theta }+{\cos }^2\mathrm{\theta }-1=0$

As  cos  0  = 1  satisfied the  equation.

$\therefore (2\cos \mathrm{\theta }-1)\left(3{\cos }^2\mathrm{\theta }+2\cos \mathrm{\theta }+1\right)=0$

$\begin{array}{l}\Rightarrow \cos \mathrm{\theta }=\frac{1}{2}\text{ (other values of }\cos \mathrm{\theta }\text{ are imaginary) }\\ \therefore \mathrm{\theta }=2\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{I}\end{array}$