If 7sin²θ + 4cos²θ = 5 and θ lies in the first quadrant, then what is the value of $$\frac{{\sqrt 3 \sec \theta + \tan \theta }}{{\sqrt 2 \cot \theta - \sqrt 3 \cos \theta }}?$$

Solution(By Examveda Team)

$$\eqalign{ & 7{\sin ^2}\theta + 4{\cos ^2}\theta = 5 \cr & 3{\sin ^2}\theta + 4{\sin ^2}\theta + 4{\cos ^2}\theta = 5 \cr & 3{\sin ^2}\theta + 4 = 5 \cr & 3{\sin ^2}\theta = 5 - 4 \cr & 3{\sin ^2}\theta = 1 \cr & {\sin ^2}\theta = \frac{1}{3} \cr & \sin \theta = \frac{1}{{\sqrt 3 }} = \frac{P}{H} \cr & B = \sqrt {{{\left( {\sqrt 3 } \right)}^2} - {1^2}} = \sqrt 2 \cr & \Rightarrow \frac{{\sqrt 3 \sec \theta + \tan \theta }}{{\sqrt 2 \cot \theta - \sqrt 3 \cos \theta }} \cr & = \frac{{\sqrt 3 \times \frac{{\sqrt 3 }}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}}}{{\sqrt 2 \times \frac{{\sqrt 2 }}{1} - \sqrt 3 \times \frac{{\sqrt 2 }}{{\sqrt 3 }}}} \cr & = \frac{{\frac{4}{{\sqrt 2 }}}}{{2 - \sqrt 2 }} \cr & = \frac{{2\sqrt 2 }}{{2 - \sqrt 2 }} \cr & = \frac{{2\sqrt 2 }}{{2 - \sqrt 2 }} \times \frac{{2 + \sqrt 2 }}{{2 + \sqrt 2 }} \cr & = \frac{{2\sqrt 2 \left( {2 + \sqrt 2 } \right)}}{{4 - 2}} \cr & = 2\left( {\sqrt 2 + 1} \right) \cr} $$