If cosec²θ + cot²θ = 3\(\frac{1}{2}\), 0° < θ < 90°, than (cosθ + sinθ) is equal to:

\(\frac{\sqrt {5}+2}{ 3}\)

cosec²θ + cot²θ = 3\(\frac{1}{2}\)

1+cot²θ + cot²θ = 3\(\frac{1}{2}\)

2cot²θ = \(\frac{7}{2}\) - 1

cot²θ =\(\frac{5}{4}\)

cotθ =\(\frac{\sqrt {5}}{2}\)=\(\frac{B}{P}\)

H=\(\sqrt {(\sqrt {5})^2+(2)^2}\) = 3

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