If $$\sin \theta = \frac{a}{{\sqrt {{a^2} + {b^2}} }},$$    0° < θ < 90°, then the value of secθ + tanθ is:

A. $$\frac{{\sqrt {{a^2} + {b^2}} + a}}{b}$$

B. $$\frac{{\sqrt {{a^2} + {b^2}} + b}}{{2a}}$$

C. $$\frac{{\sqrt {{a^2} + {b^2}} + a}}{{2b}}$$

D. $$\frac{{\sqrt {{a^2} + {b^2}} + b}}{a}$$

Answer: Option A

Solution(By Examveda Team)

$$\sin \theta = \frac{a}{{\sqrt {{a^2} + {b^2}} }}$$

$$\eqalign{ & \sec \theta + \tan \theta \cr & = \frac{{\sqrt {{a^2} + {b^2}} }}{b} + \frac{a}{b} \cr & = \frac{{\sqrt {{a^2} + {b^2}} + a}}{b} \cr} $$