tana = 3/4.
=>cota=1/ tana =1:3/4=4/3
sina /cosa =tana
=> sina =tana .cosa =3/4. cosa
lại có sin^2(a)+cos^2(a)=1
<=>9/16cos^2(a)+cos^2=1
<=>25/16cos^2(a)=1
<=>cos^2(a)=16/25
=>[cosa =4/5=>sina =3/5
    [cosa =-4/5=> sina =-2/5

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

Mà \(\cos a=\dfrac{1}{5}+\sin a\)

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

Ta có \(\sin A=1,4-\cos A\)

Thế vào \(\sin^2A+\cos^2A=1\)ta được

\(25\cos^2A-35\cos A+12=0\)

\(\Leftrightarrow\orbr{\begin{cases}\cos A=0,8\\\cos A=0,6\end{cases}\Rightarrow\orbr{\begin{cases}\sin A=0,6\\\sin A=0,8\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}\cot A=\frac{4}{3}\\\cot A=\frac{3}{5}\end{cases}}\)

giả sử tam giác ABC vuông tại A

đặt Ab=c; AC=b; BC=a, \(\widehat{B}\)=A

ta có:

\(sinA+cosA=\frac{b}{a}+\frac{c}{a}=\frac{b+c}{a}=\frac{7}{5}\)

=>b+c=7

=>(b+c)²=b²+2bc+c²=49

=>\(sin^2A+cos^2A=\left(\frac{b}{a}\right)^2+\left(\frac{c}{a}\right)^2=\frac{b^2+c^2}{a^2}=\frac{a^2}{a^2}=\frac{25}{25}\)

=>b²+c²=25

ta có:

(b+c)²-b²-c²=49-25

2bc=24

bc=12

ta có: b.c=12; b+c=7

=> 3.4=4.3=1.12=12.1=2.6=6.2

mà b+c=7=> b=4,c=3 hoặc b=3,c=4

=> cot A= 4/3 hoặc 3/4

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