If $\frac{{\sin }^3\mathrm{\theta }}{\sin (2\mathrm{\theta }+\mathrm{\alpha })}=\frac{{\cos }^3\mathrm{\theta }}{\cos (2\mathrm{\theta }+\mathrm{\alpha })}\text{ and }\tan 2\mathrm{\theta }=\mathrm{\lambda tan}(3\mathrm{\theta }+\mathrm{\alpha })$ then the value of $\mathrm{\lambda }$ is

 We have
$\begin{array}{l}\frac{{\sin }^3\mathrm{\theta }}{\sin (2\mathrm{\theta }+\mathrm{\alpha })}=\frac{{\cos }^3\mathrm{\theta }}{\cos (2\mathrm{\theta }+\mathrm{\alpha })}=\mathrm{k}\left(\mathrm{let}\right)\\ \Rightarrow \frac{{\sin }^4\mathrm{\theta }}{\sin \mathrm{\theta sin}(2\mathrm{\theta }+\mathrm{\alpha })}=\frac{{\cos }^4\mathrm{\theta }}{\cos \mathrm{\theta cos}(2\mathrm{\theta }+\mathrm{\alpha })}=\mathrm{k}\\ \Rightarrow {\cos }^4\mathrm{\theta }-{\sin }^4\mathrm{\theta }=\mathrm{k}[\cos \mathrm{\theta cos}(2\mathrm{\theta }+\mathrm{\alpha })-\sin \mathrm{\theta sin}(2\mathrm{\theta }+\mathrm{\alpha }\left)\right]\\ \Rightarrow \cos 2\mathrm{\theta }=\mathrm{kcos}(3\mathrm{\theta }+\mathrm{\alpha })-----\left(\mathrm{i}\right)\end{array}$

Also,
$\begin{array}{l}\frac{{\sin }^3\mathrm{\theta cos}\mathrm{\theta }}{\sin (2\mathrm{\theta }+\mathrm{\alpha })\cos \mathrm{\theta }}=\frac{\sin {\mathrm{\theta cos}}^3\mathrm{\theta }}{\sin \mathrm{\theta cos}(2\mathrm{\theta }+\mathrm{\alpha })}=\mathrm{k}\\ \Rightarrow {\sin }^3\mathrm{\theta cos}\mathrm{\theta }+\sin {\mathrm{\theta cos}}^3\mathrm{\theta }\\ =\mathrm{k}(\sin (2\mathrm{\theta }+\mathrm{\alpha })\cos \mathrm{\theta }+\sin \mathrm{\theta cos}(2\mathrm{\theta }+\mathrm{\alpha }\left)\right)\\ \Rightarrow \sin \mathrm{\theta cos}\mathrm{\theta }\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)=\mathrm{ksin}(3\mathrm{\theta }+\mathrm{\alpha })\\ \Rightarrow \sin 2\mathrm{\theta }=2\mathrm{ksin}(3\mathrm{\theta }+\mathrm{\alpha })---\left(\mathrm{ii}\right)\end{array}$

From (1) and (2), we get 

$\tan 2\mathrm{\theta }=2\tan (3\mathrm{\theta }+\mathrm{\alpha })$