If sec2θ + tan2θ = 3 1/2 , 0° < θ < 90°, than (cosθ + sinθ) is equal…

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 =\(\frac{P+B}{H}\) = \(\frac{\sqrt {5}+2}{3}\)