$\cos\alpha=\dfrac{3}{4}\\↔\cos^2\alpha=\dfrac{9}{16}$
mà $\sin^2\alpha+\cos^2\alpha=1$
$→\sin^2\alpha=\dfrac{7}{16}$
$↔\sin\alpha=\dfrac{\sqrt 7}{4}$ (vì $\alpha$ là góc nhọn)
$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac{\sqrt 7}{4}}{\dfrac{3}{4}}=\dfrac{\sqrt 7}{3}$
$\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{\dfrac{3}{4}}{\dfrac{\sqrt 7}{4}}=\dfrac{3\sqrt 7}{7}$
Vậy $\sin\alpha=\dfrac{\sqrt 7}{4},\,\tan\alpha=\dfrac{\sqrt 7}{3},\,\cot\alpha=\dfrac{3\sqrt 7}{7}$
