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

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: \(\tan a=\frac{\sin a}{cosa}\)

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

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

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

Mà \(\cos a=\dfrac{1}{5}+\sin a\)

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

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