If sec²θ +tan²θ = 3$\frac{1}{2}$, 0^o < θ < 90^o, then (cosθ +sinθ) is equal to 

$\frac{2+\sqrt{5}}{3}$

sec²θ + tan²θ = 3\(\frac{1}{2}\)   -----(1)

and we know , sec²θ - tan²θ = 1     ------(2)

Adding 1 & 2

2sec²θ = \(\frac{7}{2}\) + 1

sec²θ = \(\frac{9}{4}\)

secθ = \(\frac{3}{2}\)

H = 3 , B = 2

P² + B² = H²

P² = 9 - 4

P = √5

Now , (cosθ +sinθ)

= \(\frac{2}{3}\) + \(\frac{√5}{3}\)

= \(\frac{2 + √5}{3}\)