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

\(\frac{\sqrt {7}+2}{ √11}\) \(\frac{\sqrt {7}+2}{ √13}\) \(\frac{\sqrt {7}+4}{ √11}\) \(\frac{\sqrt {3}+2}{ √11}\)

\(\frac{\sqrt {7}+2}{ √11}\)

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