If sinθ = √3cosθ, 0°< θ < 90°, then the value of 2sin²θ + 6sec²θ + sinθ secθ + cosecθ is

Solution (By Examveda Team)

$$\eqalign{ & \sin \theta = \sqrt 3 \cos \theta \cr & \tan \theta = \sqrt 3 \cr & \tan \theta = \tan {60^ \circ } \cr & \theta = {60^ \circ } \cr & 2{\sin ^2}{60^ \circ } + {\sec ^2}{60^ \circ } + \tan {60^ \circ } + {\text{cosec}}\,{60^ \circ } \cr & = 2 \times \frac{3}{4} + 4 + \sqrt 3 + \frac{2}{{\sqrt 3 }} \cr & = \frac{3}{2} + 4 + \sqrt 3 + \frac{2}{{\sqrt 3 }} \cr & = \frac{{11}}{2} + \frac{5}{{\sqrt 3 }} \cr & = \frac{{11\sqrt 3 + 10}}{{2\sqrt 3 }} \cr & = \frac{{33 + 10\sqrt 3 }}{6} \cr} $$