If $$cos^4θ - sin^4θ$$ = 1/3, then the value of $$tan^2θ$$ is

Given : $$cos^4\theta-sin^4\theta=\frac{1}{3}$$

=> $$(cos^2\theta-sin^2\theta)(cos^2\theta+sin^2\theta)=\frac{1}{3}$$

=> $$(cos^2\theta-sin^2\theta)(1)=\frac{1}{3}$$

=> $$cos^2\theta-sin^2\theta=\frac{1}{3}$$ -------------(i)

Also, $$cos^2\theta+sin^2\theta=1$$ ----------(ii)

Adding equations (i) and (ii),

=> $$2cos^2\theta=1+\frac{1}{3}=\frac{4}{3}$$

=> $$cos^2\theta=\frac{2}{3}$$

Similarly, $$sin^2\theta=\frac{1}{3}$$

$$\therefore$$ $$tan^2\theta=\frac{sin^2\theta}{cos^2\theta}$$

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

=> Ans - (A)