If 2(cos2θ - sin2θ) = 1, (θ is positive acute angle), then cotθ is equal to?

Answer: Option D

$$\begin{array}{rl}& \text{2}\left(\text{co}{\text{s}}^2\theta -{\sin }^2\theta \right)=1\\ & \Rightarrow \text{co}{\text{s}}^2\theta -\left(1-\text{co}{\text{s}}^2\theta \right)=\frac{1}{2}\\ & \Rightarrow 2\text{co}{\text{s}}^2\theta =1+\frac{1}{2}\\ & \Rightarrow 2\text{co}{\text{s}}^2\theta =\frac{3}{2}\\ & \Rightarrow \text{co}{\text{s}}^2\theta =\frac{3}{4}\\ & \Rightarrow \text{se}{\text{c}}^2\theta =\frac{4}{3}\\ & \Rightarrow 1+\text{ta}{\text{n}}^2\theta =\frac{4}{3}\\ & \Rightarrow \text{ta}{\text{n}}^2\theta =\frac{4}{3}-1\\ & \Rightarrow \text{ta}{\text{n}}^2\theta =\frac{1}{3}\\ & \Rightarrow \tan \theta =\frac{1}{\sqrt{3}}\\ & \Rightarrow \cot \theta =\sqrt{3}\\ & \\ & \mathbf{Alternate}:\\ & \text{2}\left(\text{co}{\text{s}}^2\theta -{\sin }^2\theta \right)=1\\ & \Rightarrow \text{co}{\text{s}}^2\theta -\left(\text{1}-\text{co}{\text{s}}^2\theta \right)=\frac{1}{2}\\ & \Rightarrow 2\text{co}{\text{s}}^2\theta =1+\frac{1}{2}\\ & \Rightarrow \text{co}{\text{s}}^2\theta =\frac{3}{2}\times \frac{1}{2}\\ & \Rightarrow \text{co}{\text{s}}^2\theta =\frac{3}{4}\\ & \Rightarrow \cos \theta =\frac{\sqrt{3}}{2}\left[\cos {30}^{\circ }=\frac{\sqrt{3}}{2}\right]\\ & \Rightarrow \theta =30^{\circ }\\ & \text{Hence, }\\ & \text{cot}\theta =\text{cot3}{0}^{\circ }=\sqrt{3}\end{array}$$