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}}}$

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

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