\(\cos a-\sin a=\dfrac{1}{5}\\ \Leftrightarrow\left(\cos a-\sin a\right)^2=\dfrac{1}{25}\\ \Leftrightarrow1-2\sin a\cos a=\dfrac{1}{25}\\ \Leftrightarrow2\sin a\cos a=\dfrac{24}{25}\)

Mà \(\cos a=\dfrac{1}{5}+\sin a\)

\(\Leftrightarrow2\sin a\left(\dfrac{1}{5}+\sin a\right)=\dfrac{24}{25}\\ \Leftrightarrow\dfrac{2}{5}\sin a+2\sin^2a-\dfrac{24}{25}=0\\ \Leftrightarrow\left[{}\begin{matrix}\sin a=\dfrac{3}{5}\\\sin a=-\dfrac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\cos a=\dfrac{4}{5}\\\cos a=-\dfrac{3}{5}\end{matrix}\right.\\ \Leftrightarrow\cot a=\dfrac{4}{5}\cdot\dfrac{5}{3}=\dfrac{4}{3}\)

Lung tung hả

Hai câu cuối ko thấu rỏ bạn ơi

Bài 2: 

\(\cos\widehat{A}=\dfrac{3\sqrt{39}}{20}\)

\(\tan\widehat{A}=\dfrac{7}{20}:\dfrac{3\sqrt{39}}{20}=\dfrac{7}{3\sqrt{39}}=\dfrac{7\sqrt{39}}{117}\)

\(\cot\widehat{A}=\dfrac{3\sqrt{39}}{7}\)

a: Ta có: \(\sin^2a+cos^2a=1\)

=>\(cos^2a=1-0,8^2=1-0,64=0,36=0,6^2\)

=>cosa=0,6

\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)

\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)

b: Ta có: \(\sin^2a+cos^2a=1\)

=>\(\sin^2a=1-0,6^2=1-0,36=0,64=0,8^2\)

=>sin a=0,8

\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)

\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)

c: \(\tan a\cdot\cot a=1\)

=>\(\cot a=\frac13\)

Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)

=>\(\frac{1}{cos^2a}=1+3^2=10\)

=>\(cos^2a=\frac{1}{10}\)

=>\(cosa=\frac{1}{\sqrt{10}}\)

Ta có: \(\tan a=\frac{\sin a}{cosa}\)

=>\(\sin a=\tan a\cdot cosa\)

\(=3\cdot\frac{1}{\sqrt{10}}=\frac{3}{\sqrt{10}}\)

d: \(\tan a\cdot\cot a=1\)

=>\(\tan a\cdot2=1\)

=>\(\tan a=\frac12\)

Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)

=>\(\frac{1}{cos^2a}=1+\left(\frac12\right)^2=\frac54\)

=>\(cos^2a=\frac45\)

=>\(cosa=\frac{2}{\sqrt5}\)

Ta có: \(\tan a=\frac{\sin a}{cosa}\)

=>\(\sin a=cosa\cdot\tan a=\frac{2}{\sqrt5}\cdot\frac12=\frac{1}{\sqrt5}\)

a) Có: `1+tan^2a=1/(cos^2a)`

`<=> 1+(3/5)^2=1/(cos^2a)`

`=> cosa=\sqrt10/4`

`=> sina = \sqrt(1-cos^2a) = \sqrt6/4`

b) Có: `sin^2a + cos^2a=1`

`<=> sin^2a + (1/4)^2=1`

`=> sina=\sqrt15/4`

`=> tana = (sina)/(cosa) = \sqrt15`

Má ơi,tính sai:

a)\(\left[{}\begin{matrix}cos\alpha=\dfrac{5\sqrt{34}}{34}\\cos\alpha=\dfrac{-5\sqrt{34}}{34}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}sin\alpha=cos\alpha.tan\alpha=\dfrac{3\sqrt{34}}{34}\\sin\alpha=cos\alpha.tan\alpha=\dfrac{-3\sqrt{34}}{34}\end{matrix}\right.\)

b)\(\left[{}\begin{matrix}sin\alpha=\dfrac{\sqrt{15}}{4}\\sin\alpha=\dfrac{-\sqrt{15}}{4}\end{matrix}\right.\)\(\left[{}\begin{matrix}tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\sqrt{15}\\tatn\alpha=-\sqrt{15}\end{matrix}\right.\)

\(\sin^2\widehat{A}+\cos^2\widehat{A}=1\Leftrightarrow\cos^2\widehat{A}=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\\ \Leftrightarrow\cos\widehat{A}=\dfrac{4}{5}\\ \tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}=\dfrac{3}{4}\\ \Rightarrow\cot\widehat{A}=\dfrac{1}{\tan\widehat{A}}=\dfrac{4}{3}\)

Ta có: \(\sin^2A+cos^2A=1\)

=>\(cos^2A=1-\left(\frac35\right)^2=1-\frac{9}{25}=\frac{16}{25}\)

=>\(cosA=\frac45\)

tan A=sin A:cosA

\(=\frac35:\frac45=\frac35\times\frac54=\frac34\)

\(\cot A=1:\frac34=\frac43\)