If tan4θ + tan2θ = 1, then the value of cos4θ + cos2θ is?

Answer: Option C

$$\begin{array}{rl}& \text{ta}{\text{n}}^4\theta +\text{ta}{\text{n}}^2\theta =1......(\text{i})\\ & \because {\sec }^2\theta -\text{ta}{\text{n}}^2\theta =1\\ & \Rightarrow \text{ta}{\text{n}}^2\theta \left(1+\text{ta}{\text{n}}^2\theta \right)=1\\ & \Rightarrow \text{ta}{\text{n}}^2\theta \left({\sec }^2\theta \right)=1\\ & \Rightarrow \text{ta}{\text{n}}^2\theta =\frac{1}{{\sec }^2\theta }\\ & \Rightarrow \text{ta}{\text{n}}^2\theta =\text{co}{\text{s}}^2\theta \\ & \because \text{ co}{\text{s}}^4\theta +\text{co}{\text{s}}^2\theta \\ & ={\left(\text{co}{\text{s}}^2\theta \right)}^2+\text{co}{\text{s}}^2\theta \\ & ={\left(\text{ta}{\text{n}}^2\theta \right)}^2+\text{ta}{\text{n}}^2\theta \\ & =\text{ta}{\text{n}}^4\theta +\text{ta}{\text{n}}^2\theta \\ & =1\text{ from equation }\left(\text{i}\right)\end{array}$$