If cos2θ - sin2θ = 1/3 , where 0 ≤ θ ≤ /2 , then the value of cos4θ - sin4θ is?

Answer: Option A

$$\begin{array}{rl}& \text{co}{\text{s}}^2\theta -{\sin }^2\theta =\frac{1}{3}\left(\text{Given}\right)\\ & \text{co}{\text{s}}^2\theta +{\sin }^2\theta =1\left(\text{Property}\right)\\ & \left(\text{co}{\text{s}}^2\theta -{\sin }^2\theta \right)\left(\text{co}{\text{s}}^2\theta +{\sin }^2\theta \right)=\frac{1}{3}\times 1\\ & \text{co}{\text{s}}^4\theta -{\sin }^4\theta =\frac{1}{3}\\ & \therefore \left(\left(a^2+b^2\right)\left(a^2-b^2\right)=a^4-b^4\right)\end{array}$$