If $$\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta - 3\cos \theta + 2}} = 1,\,\theta $$     lies in the first quadrant, then the value of $$\frac{{{{\tan }^2}\frac{\theta }{2} + {{\sin }^2}\frac{\theta }{2}}}{{\tan \theta + \sin \theta }}$$    is:

Solution(By Examveda Team)

$$\eqalign{ & \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta - 3\cos \theta + 2}} = 1 \cr & {\sin ^2}\theta = {\cos ^2}\theta - 3\cos \theta + 2 \cr & 1 - {\cos ^2}\theta = {\cos ^2}\theta - 3\cos \theta + 2 \cr & 0 = 2{\cos ^2}\theta - 3\cos \theta + 1 \cr & 2{\cos ^2}\theta - 3\cos \theta + 1 = 0 \cr & 2{\cos ^2}\theta - 2\cos \theta - \cos \theta + 1 = 0 \cr & 2\cos \theta \left( {\cos \theta - 1} \right) - 1\left( {\cos \theta - 1} \right) = 0 \cr & \left( {2\cos \theta - 1} \right)\left( {\cos \theta - 1} \right) = 0 \cr & \cos \theta = \frac{1}{2} = \cos {60^ \circ } \cr & \cos \theta = 1 = \cos {0^ \circ } \cr & \Rightarrow \frac{{{{\tan }^2}\frac{\theta }{2} + {{\sin }^2}\frac{\theta }{2}}}{{\tan \theta + \sin \theta }} \cr & = \frac{{{{\tan }^2}\frac{{{{60}^ \circ }}}{2} + {{\sin }^2}\frac{{{{60}^ \circ }}}{2}}}{{\tan {{60}^ \circ } + \sin {{60}^ \circ }}} \cr & = \frac{{{{\tan }^2}{{30}^ \circ } + {{\sin }^2}{{30}^ \circ }}}{{\tan {{60}^ \circ } + \sin {{60}^ \circ }}} \cr & = \frac{{{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}}}{{\sqrt 3 + \frac{{\sqrt 3 }}{2}}} \cr & = \frac{{\frac{1}{3} + \frac{1}{4}}}{{\frac{{2\sqrt 3 + \sqrt 3 }}{2}}} \cr & = \frac{{\frac{{4 + 3}}{{12}}}}{{\frac{{3\sqrt 3 }}{2}}} \cr & = \frac{{7 \times 2}}{{12 \times 3\sqrt 3 }} \cr & = \frac{7}{{18\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{7\sqrt 3 }}{{54}} \cr} $$