If $$\frac{{\sin \theta }}{{1 + \cos \theta }} + \frac{{1 + \cos \theta }}{{\sin \theta }} = \frac{1}{{\sqrt 3 }},$$ 0° < θ < 90° then the value of (tanθ + secθ)-1 is?
A. 3 - √2
B. 2 + √3
C. 2 - √3
D. 3 + √2
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \frac{{\sin \theta }}{{1 + \cos \theta }} + \frac{{1 + \cos \theta }}{{\sin \theta }} = \frac{1}{{\sqrt 3 }} \cr & \frac{{{{\sin }^2}\theta + {{\left( {1 + \cos \theta } \right)}^2}}}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{1}{{\sqrt 3 }} \cr & \frac{{{{\sin }^2}\theta + 1 + {{\cos }^2}\theta + 2\cos \theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{1}{{\sqrt 3 }} \cr & \frac{{{{\sin }^2}\theta + {{\cos }^2}\theta + 1 + 2\cos \theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{1}{{\sqrt 3 }} \cr & \frac{{2\left( {1 + \cos \theta } \right)}}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{2}{{\sqrt 3 }} \cr & \sin \theta = \frac{{\sqrt 3 }}{2} \cr & \theta = {60^ \circ } \cr & \sec \theta - \tan \theta \cr & = \sec {60^ \circ } - \tan {60^ \circ } \cr & = 2 - \sqrt 3 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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