Ta có: \(\dfrac{AB}{BC}=\dfrac{1}{2}\)

\(\Rightarrow cos\alpha=\dfrac{1}{2}\)

Mà: \(sin^2\alpha+cos^2\alpha=1\)

\(\Rightarrow sin^2\alpha=1-cos^2\alpha\)

\(\Rightarrow sin^2\alpha=1-\dfrac{1}{2}\)

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

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

Mà: \(tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)

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

\(\Rightarrow tan\alpha=\sqrt{2}\)

Bài 1:

a) Ta có:

\(tanB=\dfrac{AC}{AB}\Rightarrow\dfrac{AC}{AB}=\dfrac{5}{2}\)

\(\Rightarrow AC=\dfrac{AB\cdot5}{2}=\dfrac{6\cdot5}{2}=15\)  

b) Áp dụng Py-ta-go ta có: 

\(BC^2=AB^2+AC^2=6^2+15^2=261\)

\(\Rightarrow BC=\sqrt{261}=3\sqrt{29}\)

Bài 2: 

\(\left\{{}\begin{matrix}sinM=sin40^o\approx0,64\Rightarrow cosN\approx0,64\\cosM=cos40^o\approx0,77\Rightarrow sinN\approx0,77\\tanM=tan40^o\approx0,84\Rightarrow cotN\approx0,84\\cotM=cot40^o\approx1,19\Rightarrow tanN\approx1,19\end{matrix}\right.\)

Bạn sử dụng 2 CT này để tìm nhé
\(1+\tan^2\alpha=\frac{1}{\cos^2\alpha}\)
\(\Rightarrow1+\tan^2\alpha=\frac{1}{1-\sin^2\alpha}\)

Cạnh huyền là: 8²+15²=17²

\(\Rightarrow\)\(\sin\)\(\alpha\)=\(\frac{8}{17}\)

\(\Rightarrow\)\(\cos\)\(\alpha\)=\(\frac{15}{17}\)

1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)

\(C=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha\)

\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)

\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)

1: \(cota=\sqrt{5}\)

=>\(cosa=\sqrt{5}\cdot sina\)

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

=>\(\dfrac{1}{sin^2a}=1+5=6\)

=>\(sin^2a=\dfrac{1}{6}\)

\(C=sin^2a-sina\cdot\sqrt{5}\cdot sina+\left(\sqrt{5}\cdot sina\right)^2\)

\(=sin^2a\left(1-\sqrt{5}+5\right)=\dfrac{1}{6}\cdot\left(6-\sqrt{5}\right)\)

2: tan a=3

=>sin a=3*cosa 

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

=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)

\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)

\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)

\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)