If $\sin \mathrm{\theta },\tan \mathrm{\theta },\cos \mathrm{\theta }$ are in G.P. then $4{\sin }^2\mathrm{\theta }-3{\sin }^4\mathrm{\theta }+{\sin }^6\mathrm{\theta }=$

 Given that ${\tan }^2\mathrm{\theta }=\sin \mathrm{\theta cos}\mathrm{\theta }$

$\text{∴}\sin \mathrm{\theta }={\cos }^3\mathrm{\theta }$

Also, given expression can be rewritten as 

$\begin{array}{l}{\sin }^6\mathrm{\theta }-3{\sin }^4\mathrm{\theta }+3{\sin }^2\mathrm{\theta }-1+1+{\sin }^2\mathrm{\theta }\\ ={\left({\sin }^2\mathrm{\theta }-1\right)}^3+1+{\sin }^2\mathrm{\theta }\\ =-{\left(1-{\sin }^2\mathrm{\theta }\right)}^3+1+{\sin }^2\mathrm{\theta }\\ =-{\cos }^6\mathrm{\theta }+1+{\sin }^2\mathrm{\theta }\\ =-{\sin }^2\mathrm{\theta }+1+{\sin }^2\mathrm{\theta }=1\end{array}$