If 0° < θ < 90° and cos2θ = 3(cot2θ - cos2θ) then the value of ( 1/2 + )^ - 1 is:

Answer: Option B

$$\begin{array}{rl}& {\cos }^2\theta =3\left({\cot }^2\theta -{\cos }^2\theta \right)\\ & {\cos }^2\theta =3{\cos }^2\theta \left(\frac{1}{{\sin }^2\theta }-1\right)\\ & 1=3\left(\text{cose}{\text{c}}^2\theta -1\right)\\ & \frac{1}{3}+1=\text{cose}{\text{c}}^2\theta \\ & \frac{2}{\sqrt{3}}=\text{cosec}\theta \\ & \text{cosec }60^{\circ }=\text{cosec}\theta \\ & \theta =60^{\circ }\\ & \Rightarrow {\left(\frac{1}{2}\sec \theta +\sin \theta \right)}^{-1}\\ & ={\left(\frac{1}{2}\sec {60}^{\circ }+\sin {60}^{\circ }\right)}^{-1}\\ & ={\left(\frac{1}{2}\times 2+\frac{\sqrt{3}}{2}\right)}^{-1}\\ & ={\left(\frac{2+\sqrt{3}}{2}\right)}^{-1}\\ & =\frac{2}{2+\sqrt{3}}\\ & =2\left(2-\sqrt{3}\right)\end{array}$$