Kiến thức sử dụng:

$0<\alpha<\frac{\pi}{2}\implies\cos \alpha>0$

$\sin^2x+\cos^2x=1$

$\tan x=\frac{\sin x}{\cos x}$

$\cot x=\frac{1}{\tan x}\,(\tan x\neq0)$

$\sin2x=2\sin x\cos x$

$\cos2x=\cos^2x-\sin^2x$

$\tan2x=\frac{2\tan x}{1-\tan^2x}$

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Lời giải

$\sin^2x+\cos^2x=1$

$\Longleftrightarrow\Big(\frac{2}{\sqrt{5}}\Big)^2+\cos^2x=1$

$\Longleftrightarrow\cos^2x=\frac{1}{5}$

$\implies\cos x=\frac{1}{\sqrt{5}}$ (Vì $0<\alpha<\frac{\pi}{2}$ nên $\cos\alpha>0$)

$\tan x=\frac{\sin x}{\cos x}=\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}=2$

$\cot x=\frac{1}{\tan x}=\frac{1}{2}$

*Bonus:

$\sin2x=2\sin x\cos x=\frac{2}{\sqrt{5}}.\frac{1}{\sqrt{5}}=\frac{2}{5}$

$\cos2x=\cos^2x-\sin^2x=\frac{1}{5}-\Big(\frac{2}{\sqrt{5}}\Big)^2=-\frac{3}{5}$

$\tan2x=\frac{2\tan x}{1-\tan^2x}=\frac{2.2}{1-2^2}=-\frac{4}{3}$

$\cot2x=\frac{1}{\tan2x}=\frac{1}{-\frac{4}{3}}=-\frac{3}{4}$