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

Solution (By Examveda Team)

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