Phương pháp giải:

Sử dụng công thức lượng giác: \(\left\{ \begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\\tan \alpha .\cot \alpha = 1\\1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\end{array} \right..\)

Lời giải chi tiết:

Ta có: \(0 < \alpha < {90^0}\) \( \Rightarrow \left\{ \begin{array}{l}\sin \alpha > 0\\\cos \alpha > 0\\\tan \alpha > 0\\\cot \alpha > 0\end{array} \right..\)

\(\sin \alpha = \frac{2}{3}\)

*\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)\( \Leftrightarrow {\left( {\frac{2}{3}} \right)^2} + {\cos ^2}\alpha = 1\)\( \Leftrightarrow {\cos ^2}\alpha = 1 - \frac{4}{9} = \frac{5}{9}\)\( \Rightarrow \cos \alpha = \frac{{\sqrt 5 }}{3}\)

*\(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{2}{3}:\frac{{\sqrt 5 }}{3} = \frac{{2\sqrt 5 }}{5}\)

*\(\cot \alpha = \frac{1}{{\tan \alpha }} = 1:\frac{{2\sqrt 5 }}{5} = \frac{{\sqrt 5 }}{2}\)

Chọn C.

Phương pháp giải:

Sử dụng công thức lượng giác: \(\left\{ \begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\\tan \alpha .\cot \alpha = 1\\1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\end{array} \right..\)

Lời giải chi tiết:

Ta có: \(0 < \alpha < {90^0}\) \( \Rightarrow \left\{ \begin{array}{l}\sin \alpha > 0\\\cos \alpha > 0\\\tan \alpha > 0\\\cot \alpha > 0\end{array} \right..\)

\(\tan \alpha = \frac{4}{3}\)

* \(\tan \alpha .\cot \alpha = 1\)\( \Leftrightarrow \cot \alpha = 1:tan\alpha = 1:\frac{4}{3} = \frac{3}{4}\)

* \(1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\)\( \Leftrightarrow 1 + {\left( {\frac{4}{3}} \right)^2} = \frac{1}{{{{\cos }^2}\alpha }}\)\( \Leftrightarrow \frac{1}{{{{\cos }^2}\alpha }} = \frac{{25}}{9}\)\( \Rightarrow {\cos ^2}\alpha = \frac{9}{{25}}\)\( \Rightarrow \cos \alpha = \frac{3}{5}\)

*\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)\( \Leftrightarrow {\left( {\frac{3}{5}} \right)^2} + {\sin ^2}\alpha = 1\)\( \Leftrightarrow {\sin ^2}\alpha = 1 - \frac{9}{{25}} = \frac{{16}}{{25}}\)\( \Rightarrow \sin \alpha = \frac{4}{5}\)

Chọn B.