The value of $$1 + \sqrt {\frac{{\cot \theta + \cos \theta }}{{\cot \theta - \cos \theta }}} ,$$ if 0° < θ < 90°, is equal to:
A. 1 - secθ + tanθ
B. 1 - secθ - tanθ
C. 1 + secθ - tanθ
D. 1 + secθ + tanθ
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & 1 + \sqrt {\frac{{\cot \theta + \cos \theta }}{{\cot \theta - \cos \theta }}} \cr & = 1 + \sqrt {\frac{{\frac{{\cos \theta }}{{\sin \theta }} + \cos \theta }}{{\frac{{\cos \theta }}{{\sin \theta }} - \cos \theta }}} \cr & = 1 + \sqrt {\frac{{\cot \theta + \sin \theta \cos \theta }}{{\cot \theta - \cos \theta \sin \theta }}} \cr & = 1 + \sqrt {\frac{{\cos \theta \left( {1 + \sin \theta } \right)}}{{\cos \theta \left( {1 - \sin \theta } \right)}}} \cr & = 1 + \sqrt {\frac{{1 + \sin \theta }}{{1 - \sin \theta }}} \cr & {\text{Rationalise,}} \cr & = 1 + \sqrt {\frac{{1 + \sin \theta }}{{1 - \sin \theta }} \times \frac{{1 + \sin \theta }}{{1 + \sin \theta }}} \cr & = 1 + \sqrt {\frac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{1 - {{\sin }^2}\theta }}} \cr & = 1 + \sqrt {{{\left( {1 + \sin \theta } \right)}^2} \times {{\sec }^2}\theta } \cr & = 1 + \left( {1 + \sin \theta } \right)\sec \theta \cr & = \sec \theta + \sin \theta \sec \theta \cr & = 1 + \sec \theta + \tan \theta \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
Join The Discussion