If $$sec^2 θ + tan^2 θ = \frac{5}{3}$$, then what is the value of tan 2θ?

Given : $$sec^2 θ + tan^2 θ = \frac{5}{3}$$ --------------(i)

Also, $$sec^2\theta-tan^2\theta=1$$ -----------(ii)

Subtracting equation (ii) from (i), we get :

=> $$2tan^2\theta=\frac{5}{3}-1=\frac{2}{3}$$

=> $$tan^2\theta=\frac{2}{3}\times\frac{1}{2}=\frac{1}{3}$$

=> $$tan\theta=\sqrt{\frac{1}{3}}$$

=> $$\theta=tan^{-1}(\frac{1}{\sqrt3})$$

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

$$\therefore$$ $$tan2\theta=tan(60^\circ)=\sqrt3$$

=> Ans - (B)