giúp tui bài này vs ạ

a) 

 Ta có: $\displaystyle \sin^2\alpha + \cos^2\alpha = 1$

    $\displaystyle \Rightarrow \cos^2\alpha = 1 - \sin^2\alpha = 1 - 0,5^2 = 0,75$

    $\displaystyle \Rightarrow \cos\alpha = \sqrt{0,75} = \frac{\sqrt{3}}{2}$ (vì $\alpha$ là góc nhọn)

$\displaystyle \tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{0,5}{\frac{\sqrt{3}}{2}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

 $\displaystyle \cot\alpha = \frac{1}{\tan\alpha} = \sqrt{3}$

b) 

 Ta có: $\displaystyle \sin^2\alpha + \cos^2\alpha = 1$

    $\displaystyle \Rightarrow \sin^2\alpha = 1 - \cos^2\alpha = 1 - 0,4^2 = 0,84$

    $\displaystyle \Rightarrow \sin\alpha = \sqrt{0,84} = \frac{\sqrt{21}}{5}$ (vì $\alpha$ là góc nhọn)

 $\displaystyle \tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\frac{\sqrt{21}}{5}}{0,4} = \frac{\frac{\sqrt{21}}{5}}{\frac{2}{5}} = \frac{\sqrt{21}}{2}$

 $\displaystyle \cot\alpha = \frac{1}{\tan\alpha} = \frac{1}{\frac{\sqrt{21}}{2}} = \frac{2}{\sqrt{21}} = \frac{2\sqrt{21}}{21}$