$\because \cos \theta, \sin \theta, \cot \theta$ are in geometric progression.
$$
\begin{aligned}
& \therefore \frac{\sin \theta}{\cos \theta}=\frac{\cot \theta}{\sin \theta} \\
& \Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{\cos \theta}{\sin ^2 \theta} \Rightarrow \sin ^3 \theta=\cos ^2 \theta.....(i)
\end{aligned}
$$
$$
\begin{aligned}
& \text { Now, } \sin ^9 \theta+\sin ^6 \theta+3 \sin ^5 \theta+\sin ^3 \theta+\sin ^2 \theta \\
& =\cos ^6 \theta+\sin ^6 \theta+3 \sin ^5 \theta+\cos ^2 \theta+\sin ^2 \theta \quad \text { [from (i)] } \\
& =\left(\cos ^2 \theta+\sin ^2 \theta\right)\left(\cos ^4 \theta+\sin ^4 \theta-\cos ^2 \theta \sin ^2 \theta\right)+3 \\
& \sin ^5 \theta+1 \\
& =\left(\cos ^2 \theta+\sin ^2 \theta\right)^2-3 \sin ^2 \theta \cos ^2 \theta+3 \sin ^5 \theta+1 \\
& =1^2-3 \sin ^5 \theta+3 \sin ^5 \theta+1=2
\end{aligned}
$$