Nếu alpha là góc nhọn và sin alpha/2 = căn (x - 1)/2x thì tan alpha bằng

* Hướng dẫn giải

Đáp án B

Ta có:

$$\begin{gathered} 0<\alpha <{90}^0\iff 0<\frac{\alpha }{2}<{45}^0\Rightarrow 0<\sin \frac{\alpha }{2}<\frac{\sqrt{2}}{2} \\ \iff 0<\sqrt{\frac{x-1}{2x}}<\frac{\sqrt{2}}{2}\iff x>0 \\ {\sin }^2\frac{\alpha }{2}+{\cos }^2\frac{\alpha }{2}=1\Rightarrow \cos \frac{\alpha }{2}=\sqrt{1-{\sin }^2\frac{\alpha }{2}} \end{gathered}$$

vì $0<\frac{\alpha }{2}<{45}^0$

$$\begin{gathered} \iff \cos \frac{\alpha }{2}=\sqrt{\frac{x+1}{2x}}\Rightarrow \tan \frac{\alpha }{2}=\sqrt{\frac{x-1}{x+1}} \\ \tan \alpha =\frac{2\tan \frac{\alpha }{2}}{1-{\tan }^2\frac{\alpha }{2}}=\frac{2\sqrt{\frac{x-1}{x+1}}}{1-\frac{x-1}{x+1}}=\sqrt{x^2-1} \end{gathered}$$