If $$\frac{{\sin \theta }}{{1 + \cos \theta }} + \frac{{1 + \cos \theta }}{{\sin \theta }} = \frac{1}{{\sqrt 3 }},$$      0° < θ < 90° then the value of (tanθ + secθ)^-1 is?

$$\eqalign{ & \frac{{\sin \theta }}{{1 + \cos \theta }} + \frac{{1 + \cos \theta }}{{\sin \theta }} = \frac{1}{{\sqrt 3 }} \cr & \frac{{{{\sin }^2}\theta + {{\left( {1 + \cos \theta } \right)}^2}}}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{1}{{\sqrt 3 }} \cr & \frac{{{{\sin }^2}\theta + 1 + {{\cos }^2}\theta + 2\cos \theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{1}{{\sqrt 3 }} \cr & \frac{{{{\sin }^2}\theta + {{\cos }^2}\theta + 1 + 2\cos \theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{1}{{\sqrt 3 }} \cr & \frac{{2\left( {1 + \cos \theta } \right)}}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{2}{{\sqrt 3 }} \cr & \sin \theta = \frac{{\sqrt 3 }}{2} \cr & \theta = {60^ \circ } \cr & \sec \theta - \tan \theta \cr & = \sec {60^ \circ } - \tan {60^ \circ } \cr & = 2 - \sqrt 3 \cr} $$