If $$\frac{{{{\cos }^2}\theta }}{{{{\cot }^2}\theta - {{\cos }^2}\theta }} = 3,$$    0° < θ < 90°, then the value of cotθ + cosecθ is:

Solution (By Examveda Team)

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