If sinθ = √3cosθ, 0°< θ < 90°, then the value of 2sin2θ + 6sec2θ + sinθ secθ + cosecθ is
A. $$\frac{{33 + 10\sqrt 3 }}{6}$$
B. $$\frac{{19 + 10\sqrt 3 }}{3}$$
C. $$\frac{{19 + 10\sqrt 3 }}{6}$$
D. $$\frac{{33 + 10\sqrt 3 }}{3}$$
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \sin \theta = \sqrt 3 \cos \theta \cr & \tan \theta = \sqrt 3 \cr & \tan \theta = \tan {60^ \circ } \cr & \theta = {60^ \circ } \cr & 2{\sin ^2}{60^ \circ } + {\sec ^2}{60^ \circ } + \tan {60^ \circ } + {\text{cosec}}\,{60^ \circ } \cr & = 2 \times \frac{3}{4} + 4 + \sqrt 3 + \frac{2}{{\sqrt 3 }} \cr & = \frac{3}{2} + 4 + \sqrt 3 + \frac{2}{{\sqrt 3 }} \cr & = \frac{{11}}{2} + \frac{5}{{\sqrt 3 }} \cr & = \frac{{11\sqrt 3 + 10}}{{2\sqrt 3 }} \cr & = \frac{{33 + 10\sqrt 3 }}{6} \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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