If the sinθ + cosθ =\(\sqrt{5}\) sin(90°-θ), then find the value of cotθ.

\(\frac{\sqrt {5}-1}{4}\) \(\frac{\sqrt {5}+1}{4}\) \(\frac{\sqrt {5}+1}{3}\) \(\frac{\sqrt {3}+\sqrt {5}}{2}\)

\(\frac{\sqrt {5}+1}{4}\)

sinθ + cosθ =\(\sqrt{5}\)cosθ    (sin(90°-θ) = cosθ) sinθ =\(\sqrt{5}\)cosθ - cosθ sinθ =cosθ(\(\sqrt{5}-1\)) \(\frac{cosθ}{sinθ}\) = \(\frac{1}{\sqrt {5}-1}\) cotθ = \(\frac{\sqrt {5}+1}{4}\)  (By rationalisation)