If $${\sqrt{6}}tan\theta=\sqrt{2}$$ and $$0^\circ<\theta<45^\circ$$, then the value of $$sin\theta+\sqrt{3}cos\theta-2tan^2\theta$$ is

Given : $${\sqrt{6}}tan\theta=\sqrt{2}$$

=> $$tan\theta=\frac{\sqrt2}{\sqrt6}=\frac{1}{\sqrt3}$$

=> $$\theta=30^\circ$$

$$\therefore$$ $$sin\theta+\sqrt{3}cos\theta-2tan^2\theta$$

= $$sin(30^\circ)+\sqrt3cos(30^\circ)-2tan^2(30^\circ)$$

= $$\frac{1}{2}+(\sqrt3\times\frac{\sqrt3}{2})-2(\frac{1}{\sqrt3})^2$$

= $$\frac{1}{2}+\frac{3}{2}-\frac{2}{3}$$

= $$\frac{3+9-4}{6}=\frac{8}{6}=\frac{4}{3}$$

=> Ans - (B)