If sinθ = a/ [√(a2 + b2), 0° < θ < 90°, then the value of secθ + tanθ is:

\(\frac{{\sqrt {{a^2}{\rm{\;}} + {\rm{\;}}{b^2}} {\rm{\;}} + {\rm{\;}}a}}{{2b}}\)

\(\frac{{\sqrt {{a^2}{\rm{\;}} + {\rm{\;}}{b^2}} {\rm{\;}} + {\rm{\;}}b}}{{2a}}\)

\(\frac{{\sqrt {{a^2}{\rm{\;}} + {\rm{\;}}{b^2}} {\rm{\;}} + {\rm{\;}}a}}{a}\)

\(\frac{{\sqrt {{a^2}{\rm{\;}} + {\rm{\;}}{b^2}} {\rm{\;}} + {\rm{\;}}a}}{b}\)

If sinθ = a/ [√(a2 + b2), 0° < θ < 90°, then the value of secθ

1.	If sinθ = a/ [√(a2 + b2), 0° < θ < 90°, then the value of secθ + tanθ is:
A.	\(\frac{{\sqrt {{a^2}{\rm{\;}} + {\rm{\;}}{b^2}} {\rm{\;}} + {\rm{\;}}a}}{{2b}}\)
B.	\(\frac{{\sqrt {{a^2}{\rm{\;}} + {\rm{\;}}{b^2}} {\rm{\;}} + {\rm{\;}}b}}{{2a}}\)
C.	\(\frac{{\sqrt {{a^2}{\rm{\;}} + {\rm{\;}}{b^2}} {\rm{\;}} + {\rm{\;}}a}}{a}\)
D.	\(\frac{{\sqrt {{a^2}{\rm{\;}} + {\rm{\;}}{b^2}} {\rm{\;}} + {\rm{\;}}a}}{b}\)
Answer» E.