cho góc nhọn a..............................

`a)` Ta có: `sin^2\alpha + cos^2\alpha = 1`
`=> sin^2\alpha = 1 - cos^2\alpha = 1-1/9 = 8/9`
`=>  sin\alpha =  = \sqrt{(8/9)} = \sqrt{8}/9 = \frac{2\sqrt{2}}{3}` (do `\alpha` là góc nhọn)
Khi đó: `tg \alpha = \frac{sin\alpha}{cos\alpha} = \frac{2\sqrt{2}}{3} : 1/3 = 2\sqrt{2}`
`b)` Ta có: ` tg^2\alpha + 1 = \frac{1}{cos^2\alpha}`
`=> cos^2\alpha = \frac{1}{tg^2\alpha + 1} = \frac{1}{2^2 + 1} = 1/5`
`=> cos\alpha = \frac{\sqrt{5}}{5}` (do `\alpha` là góc nhọn)
Khi đó: `sin\alpha = cos\alpha .tg\alpha =\frac{\sqrt{5}}{5} . 2 =  \frac{2\sqrt{5}}{5}`