If sec2θ + tan2θ = 3\(\frac{1}{2}\), 0° < θ < 90°, than (cosθ + sinθ) is equal to:

\(\frac{1+\sqrt {5}}{3}\) \(\frac{2+\sqrt {5}}{3}\) \(\frac{1+\sqrt {5}}{6}\) \(\frac{9+2\sqrt {5}}{6}\)

\(\frac{2+\sqrt {5}}{3}\)

sec2θ + tan2θ = 3\(\frac{1}{2}\) 1+tan2θ + tan2θ = 3\(\frac{1}{2}\) 2tan2θ = \(\frac{7}{2}\) - 1 tan2θ =\(\frac{5}{4}\) tanθ =\(\frac{\sqrt {5}}{2}\)=\(\frac{P}{B}\) H=\(\sqrt {(\sqrt {5})^2+(2)^2}\)=3 ⇒ cosθ + sinθ =\(\frac{P+B}{H}\) = \(\frac{\sqrt {5}+2}{3}\)