If cosec²θ + cot²θ = 4\(\frac{1}{2}\), 0° < θ < 90°, than (cosθ + sinθ) is equal to:

\(\frac{\sqrt {7}+2}{ √11}\)

cosec²θ + cot²θ = 4\(\frac{1}{2}\)

1+cot²θ + cot²θ = 4\(\frac{1}{2}\)

2cot²θ = \(\frac{9}{2}\) - 1

cot²θ =\(\frac{7}{4}\)

cotθ =\(\frac{\sqrt {7}}{2}\)=\(\frac{P}{B}\)

H=\(\sqrt {(\sqrt {7})^2+(2)^2}\) = √11

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