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

 \(\frac{\sqrt {7}+2}{ √11}\)

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 =\(\frac{P+B}{H}\) = \(\frac{\sqrt {7}+2}{ √11}\)