If 1 + cos²θ = 3sinθ cosθ, then the integral value of cotθ is $$\left( {0 < \theta < \frac{\pi }{2}} \right) = \,?$$

Solution (By Examveda Team)

$$\eqalign{ & {\text{Given,}} \cr & {\text{1}} + {\text{co}}{{\text{s}}^2}\theta = {\text{3}}\sin \theta .{\text{cos}}\theta {\text{ }}\left( {0 < \theta < \frac{\pi }{2}} \right) \cr & {\text{1}} + {\text{co}}{{\text{s}}^2}\theta = {\text{3}}\sin \theta .{\text{cos}}\theta \cr & {\text{Dividing by }}{\sin ^2}\theta {\text{ in both sides}} \cr & \Rightarrow \frac{{1 + {\text{co}}{{\text{s}}^2}\theta }}{{{{\sin }^2}\theta }} = \frac{{3\sin \theta .{\text{cos}}\theta }}{{{{\sin }^2}\theta }} \cr & \Rightarrow {\text{cose}}{{\text{c}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = 3{\text{cot}}\theta \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = 3\cot \theta \cr & \left[ {\because 1 + {\text{co}}{{\text{t}}^2}\theta = {\text{cose}}{{\text{c}}^2}\theta } \right] \cr & \Rightarrow 1 + 2{\text{co}}{{\text{t}}^2}\theta = 3\cot \theta \cr & \Rightarrow 2{\text{co}}{{\text{t}}^2}\theta = 3\cot \theta - 1 \cr & {\text{Let }}\theta = {45^ \circ } \cr & \because {\text{cot4}}{{\text{5}}^ \circ } = 1 \cr & \Rightarrow 2{\text{co}}{{\text{t}}^2}{45^ \circ } - 3{\text{cot}}{45^ \circ } + 1 = 0 \cr & \Rightarrow 2 - 3 + 1 = 0 \cr & \Rightarrow 0 = 0 \cr & {\text{Therefore cot}}\theta = {\text{cot}}{45^ \circ } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \cr} $$