If sec2θ + tan2θ = 4\(\frac{1}{2}\), 0° < θ < 90°, than (cosθ + sinθ) is equal to:

\(\frac{\sqrt {3}+2}{ √11}\)  \(\frac{\sqrt {7}+2}{ √7}\)  \(\frac{\sqrt {7}+2}{ √11}\)  \(\frac{\sqrt {7}+2}{ 9}\)

\(\frac{\sqrt {7}+2}{ √11}\)

sec2θ + tan2θ = 4\(\frac{1}{2}\) 1+tan2θ + tan2θ = 4\(\frac{1}{2}\) 2tan2θ = \(\frac{9}{2}\) - 1 tan2θ =\(\frac{7}{4}\) tanθ =\(\frac{\sqrt {7}}{2}\)=\(\frac{P}{B}\) H=\(\sqrt {(\sqrt {7})^2+(2)^2}\) = √11 ⇒ cosθ + sinθ =\(\frac{P+B}{H}\) = \(\frac{\sqrt {7}+2}{ √11}\)