$\frac{1}{\cos 290°}+\frac{1}{\sqrt{3}\sin 250°}=\lambda$ then $3{\lambda }^2=$

$\lambda =\frac{1}{\sin 20°}-\frac{1}{\sqrt{3}\cos 20°}=\frac{\sqrt{3}\cos 20°-\sin 20°}{\sqrt{3}.\cos 20°.\sin 20°}=\frac{4\sin ({60}^0-{20}^0)}{\sqrt{3}\sin {40}^0}=\frac{4}{\sqrt{3}}$