If 3tanθ = 2√3sin, 0° < θ < 90°, then the value of $$\frac{{{\text{cose}}{{\text{c}}^2}2\theta + {{\cot }^2}2\theta }}{{{{\sin }^2}\theta + {{\tan }^2}\theta }}$$    is:

Solution (By Examveda Team)

$$\eqalign{ & 3\tan \theta = 2\sqrt 3 \sin \theta \cr & \cos \theta = \frac{{\sqrt 3 }}{2} = \frac{B}{H} \cr & P = \sqrt {{3^2} + {2^2}} = \sqrt {13} \cr & \cos \theta = \cos {30^ \circ } \cr & \theta = {30^ \circ } \cr & \Rightarrow \frac{{{\text{cose}}{{\text{c}}^2}2\theta + {{\cot }^2}2\theta }}{{{{\sin }^2}\theta + {{\tan }^2}2\theta }} \cr & = \frac{{{\text{cose}}{{\text{c}}^2}{{60}^ \circ } + {{\cot }^2}{{60}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {{\tan }^2}{{60}^ \circ }}} \cr & = \frac{{\frac{4}{3} + \frac{1}{3}}}{{\frac{1}{4} + 3}} \cr & = \frac{{\frac{5}{3}}}{{\frac{{13}}{4}}} \cr & = \frac{{20}}{{39}} \cr} $$