Nếu $\alpha $ là góc nhọn và $\sin \dfrac{\alpha }{2} = \sqrt {\dfrac{{x - 1}}{{2x}}} $ thì $\cot \alpha $ bằng

Ta có: $0 < \alpha  < 90^\circ $$ \Leftrightarrow 0 < \dfrac{\alpha }{2} < 45^\circ $$ \Rightarrow 0 < \sin \dfrac{\alpha }{2} < \dfrac{{\sqrt 2 }}{2}$$ \Leftrightarrow 0 < \sqrt {\dfrac{{x - 1}}{{2x}}}  < \dfrac{{\sqrt 2 }}{2}$$ \Leftrightarrow x > 0$

${\sin ^2}\dfrac{\alpha }{2} + {\cos ^2}\dfrac{\alpha }{2} = 1$$ \Rightarrow \cos \dfrac{\alpha }{2} = \sqrt {1 - {{\sin }^2}\dfrac{\alpha }{2}} $, vì $0 < \dfrac{\alpha }{2} < 45^\circ $

$ \Leftrightarrow \cos \dfrac{\alpha }{2} = \sqrt {\dfrac{{x + 1}}{{2x}}} $$ \Rightarrow \tan \dfrac{\alpha }{2} = \sqrt {\dfrac{{x - 1}}{{x + 1}}} $

$tan\alpha  = \dfrac{{2\tan \dfrac{\alpha }{2}}}{{1 - {{\tan }^2}\dfrac{\alpha }{2}}} = \dfrac{{2\sqrt {\dfrac{{x - 1}}{{x + 1}}} }}{{1 - \dfrac{{x - 1}}{{x + 1}}}} = \sqrt {{x^2} - 1} $

$ \Rightarrow \cot \alpha  = \dfrac{1}{{\tan \alpha }} = \dfrac{1}{{\sqrt {{x^2} - 1} }} = \dfrac{{\sqrt {{x^2} - 1} }}{{{x^2} - 1}}$.