The value of 2 + \(\sqrt {\frac{cot θ\;+\;cos θ}{cot θ\;-\;cos θ}}\), if 0° < θ < 90° is equal to:

2 + sec θ + tan θ

\(\sqrt {\frac{cot θ + cos θ}{cot θ - cos θ}}\) = \(\sqrt {\frac{\frac{cos θ }{sin θ}+ cos θ}{\frac{cos θ }{sin θ} - cos θ}}\)

                        = \(\sqrt {\frac{1+ sin θ}{1 - sin θ}}\)

                      = \(\sqrt {\frac{(1 + sin θ) (1+ sin θ)}{(1- sin θ) ( 1 + sin θ)}}\)

                        = \(\frac{1 + sin θ}{cos θ}\)

⇒ 2 + \(\sqrt {\frac{cot θ+ cos θ}{cot θ - cos θ}}\) = 2 + sec θ + tan θ