If 0° < θ < 90° and cos²θ = 3(cot²θ - cos²θ) then the value of $${\left( {\frac{1}{2}\sec \theta + \sin \theta } \right)^{ - 1}}$$   is:

Solution (By Examveda Team)

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