value of $2\lambda^2-1 $ in trigonometric expression if $\displaystyle \frac{\sin^3 \beta}{\sin(\beta - 3 \alpha)} = \frac{\cos^3 \beta}{\cos(\beta-3 \alpha)} = \lambda\;,$ then $2\lambda^2-1 = $
options
$(a)\; \lambda \sin \beta$
$(b)\; \lambda \cos \beta$
$(c)\; 2\lambda \cos \beta$
$(d)\; \lambda \tan \beta$
Attempt with ratio and proportion 
$\displaystyle \frac{\sin^4 \beta}{\sin \beta \sin(\beta - 3 \alpha)} = \frac{\cos^4 \beta}{\cos \beta \cos(\beta-3 \alpha)}  = \frac{\cos^4 \beta - \sin^4 \beta}{\cos \beta \cos(\beta-3 \alpha)-\sin \beta \sin(\beta - 3 \alpha)}$
so $\displaystyle \frac{\sin^4 \beta}{\sin \beta \sin(\beta - 3 \alpha)} = \frac{\cos^4 \beta}{\cos \beta \cos(\beta-3 \alpha)} = \frac{\cos 2 \beta}{\cos (2\beta-3 \alpha)}$
wan,t be able to go further ,could some help me

HINT:
Clearly we need to eliminate $\alpha$
Expand $\sin(\beta-3\alpha),\cos(\beta-3\alpha)$  to form two simultaneous equations in  $\sin(3\alpha),\cos(3\alpha)$
Solve for $\sin(3\alpha),\cos(3\alpha)$
Use $\sin^2(3\alpha)+\cos^2(3\alpha)=1$  to eliminate $\alpha$
