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

 \(\frac{\sqrt {3}+2}{ √7}\)

sec²θ + tan²θ = 2\(\frac{1}{2}\)

1+tan²θ + tan²θ = 2\(\frac{1}{2}\)

2tan²θ = \(\frac{5}{2}\) - 1

tan²θ =\(\frac{3}{4}\)

tanθ =\(\frac{\sqrt {3}}{2}\)=\(\frac{P}{B}\)

H=\(\sqrt {(\sqrt {3})^2+(2)^2}\) = √7

⇒ cosθ + sinθ =\(\frac{P+B}{H}\) = \(\frac{\sqrt {3}+2}{ √7}\)