$\sin \alpha =2$?? $\sin \alpha \in [-1;1]$ với mọi $\alpha$ mà bạn. Bạn xem lại đề.

a: \(\cos\alpha=\dfrac{1}{2}\)

\(\tan\alpha=\sqrt{3}\)

\(\cot\alpha=\dfrac{\sqrt{3}}{3}\)

Em 2k8 ms học nên k chắc

Vì 0 < \(\alpha< \dfrac{\pi}{2}\)  => sin \(\alpha>0\)

Cos \(\alpha=\dfrac{1}{3}\)  \(\Rightarrow sin\alpha=\sqrt{1-\dfrac{1}{9}}=\dfrac{2\sqrt{2}}{3}\)

tan \(\alpha=2\sqrt{2}\)  ; cot \(\alpha=\dfrac{1}{2\sqrt{2}}\)

Ta có:

\(cot\alpha\cdot tan\alpha=1\)

\(\Rightarrow cot\alpha=\dfrac{1}{tan\alpha}\)

\(\Rightarrow cota=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

Mà:

\(cot^2\alpha+1=\dfrac{1}{sin^2\alpha}\)

\(\Rightarrow sin\alpha=\sqrt{\dfrac{1}{cot^2\alpha+1}}\)

\(\Rightarrow sin\alpha=\sqrt{\dfrac{1}{\left(\dfrac{4}{3}\right)^2+1}}=\dfrac{3}{5}\) 

Lại có:

\(cos^2\alpha+sin^2\alpha=1\)

\(\Rightarrow cos\alpha=\sqrt{1-sin^2a}\)

\(\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)

\(tan\alpha=\dfrac{3}{4}\\ \Rightarrow cot\alpha=1:\dfrac{3}{4}=\dfrac{4}{3}\)

Có:

\(1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\\ \Rightarrow sin\alpha=\sqrt{1:\left(1+\left(\dfrac{4}{3}\right)^2\right)}=\dfrac{3}{5}\)

\(\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)

Có sin²a + cos²a = 1

Mà cos a = \(\dfrac{3}{4}\)

=>  sin²a + (\(\dfrac{3}{4}\))² = 1

=> sin²a + \(\dfrac{3^2}{4^2}\) = 1

=> sin²a + \(\dfrac{9}{16}\)= 1

=> sin²a = \(\dfrac{7}{16}\)

=> sin a = \(\dfrac{\sqrt{7}}{4}\)

Có tan a = \(\dfrac{\text{sin a}}{\text{cos a}}\)

Mà \(\left\{{}\begin{matrix}\text{cos a = }\dfrac{3}{4}\\\text{sin a = }\dfrac{\sqrt{7}}{4}\end{matrix}\right.\)

=> tan a = \(\dfrac{\dfrac{\sqrt{7}}{4}}{\dfrac{3}{4}}\) = \(\dfrac{\sqrt{7}}{4}\): \(\dfrac{3}{4}\) = ​\(\dfrac{\sqrt{7}}{4}\).\(\dfrac{4}{3}\) =​\(\dfrac{\sqrt{7}}{3}\)

\(A=\dfrac{\dfrac{sina}{cosa}+\dfrac{cosa}{cosa}}{\dfrac{sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{tana+1}{tana-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}\)

1+tan^2a=1/cos^2a

=>1/cos^2a=1+9/16=25/16

=>cos^2a=16/25

=>cosa=4/5 hoặc cosa=-4/5

\(1+tan^2a=\dfrac{1}{cos^2a}=1:\dfrac{1}{25}=25\)

=>tan^2a=24

=>tana=2*căn 6

\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)

\(sina=\sqrt{1-\left(\dfrac{1}{5}\right)^2}=\dfrac{2\sqrt{6}}{5}\)