If 1 + cos2θ = 3sinθ cosθ, then the integral value of cotθ is ( 0 < < /2 ) = \,?

Answer: Option B

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